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HENRY HOLT & CO., Publishers, New York. 



THE ESSENTIALS 



OF 



TRIGONOMETRY 



PLANE AND SPHERICAL 



TME Al FOl PLACE lOGARimiC Al TRIGOIITRIC TABLES 



SIMON NEWCOMB 

Professat' of Mathematics, U. S. JVavy 




I I 06'^ (P ' 



NEW YORK 
HENRY HOLT AND COMPANY 

1884 



Copyright, 1884, 

BY 

HENRY HOLT & CO. 






PEEFAOE. 



The general scope of the present work is sufficiently indi- 
cated by its title. A great number of simple exercises have 
been given, and four-place tables have been added to the 
book. These tables are all required for those who do not 
desire to become expert in computation; those who do should 
use five-place tables. 

The chapter on Surveying has been prepared by Dr. 
J. Morrison. 

The cuts of surveying instruments have been supplied by 
Messrs. W. & L. E. Gurley, of Troy, from their illustrated 
catalogue. 



TABLE OF OOE"TENTS. 



PART I. PLANE TRIGONOMETRY. 
Chapter I. Measurement of Lines and Angles. 

PAGE 

Linear Measure 1 

Angular Measure 3 

Chapter II. The Trigonometric Functions. 

Line Functions 10 

Trigonometric Functions 11 

Relations between Trigonometric Functions 16 

Functions of 45°, 30° and 18° 18 

Chapter III. Of Right Triangles. 

Fundamental Relations 22 

Solution of Right Triangles 24 

Projections of Lines 32 

Chapter IV. Trigonometric Functions op Unlimited 

Angles. 33 

Chapter V. Relations between Functions of Several Angles. 

The Addition and Subtraction Theorems 45 

The Addition Theorem for Tangents 48 

Products of Sines and Cosines 48 

Sum of Sines and Cosines 49 

Chapter VI. Solution of Triangles in General. 

Case I. Given the Angles and One Side 54 

Case II. Given Two Sides and the Angle opposite one of them . . 58 

Case III. Given the Three Sides 59 

Case IV. Given Two Sides and the Included Angle 66 

Areas of Triangles and Polygons 69 



IV CONTENTS 



PAGK 

Chapter VII. Problems in Trigonometric Computation 

AND Analysis. 72 



PART II. SPHERICAL TRIGONOMETRY. 

Ch^-TEr I. Fundamental Relations. 

Fundamental Equations 79 

Theorem of Sines 81 

Polar Triangles 82 

Transformation of the General Formulae 84 

Chapter II. Right and Quadrantal Triangles. 

Formulae for Right Triangles 88 

Napier's Rules 89 

Isosceles Triangles 92 

Chapter III. Solution op Spherical Triangles in General. 

Case I. Given the Three Sides 93 

Case II. Given the Three Angles 96 

Case III. Given Two Sides and the Included Angle 97 

Case IV. Given Two Angles and the Side between 100 

Case V. Given Two Sides and an Angle opposite one of them. . . 103 

Case VI. Given Two Angles and a Side opposite one of them 104 

Miscellaneous Exercises in Trigonometry 104 

Land Surveying 107 

Answers to Exercises 145 

Introduction to the Tables of Logarithms 153 

Three- and Four-Place Tables op Logarithms and Trigo- 
nometric Functions 167 



PAR.T I. 

PLANE TRIGONOMETRY. 



CHAPTER I. 

MEASUREMENT OF LINES AND ANGLES. 



1. Definition. Trigonometry is that branch of geom- 
etry in whioli the relations of lines and angles are treated by 
algebraic methods. 

Def. Plane trigonometry treats of lines and angles 
situated in a plane. 

Def. Spherical trigonometry treats of lines and 
angles which are not all in the same plane. 

In trigonometry the lengths of lines and the magnitudes 
of angles are expressed by algebraic symbols, or by the nu- 
merals of arithmetic. 

2. Method of expressing Quantities numerically. To ex- 
press numerically the length of a given line we assume a unit 
of length and express what multiple part or parts of the unit 
will form a line equal to the given one. 

Such multiple part or parts is the ratio of the given 
line to the unit, and the combination of figures which ex- 
presses this ratio is called a number. (Comp. Geom., § 352.) 

3. Forms of Numbers. If the line to be measured con- 
tains the unit an exact number of times, the number is an 
integer. 

If, by dividing the unit into n equal parts, the line to be 
measured contains m of these parts, its length is expressed by 



2 PLANE TRIGONOMETRY. 

the fraction — . The number is then called fractioDal or 

n 

rational. 

If the unit and the line are iJicommensuraUe, the length 
of the line may be expressed with as small an error as we 
please by an integer (which integer may be zero) plus a 
decimal fraction. The integer expresses the greatest number 
of times the line contains the unit; the fraction expresses 
the piece of the line, less than the unit, which is left over. 

To express length by a decimal fraction the unit is sup- 
posed to be diyided into 10 parts called tenths; one of these 
parts into 10 others called hundredths; the hundredth into 
10 others called thousandths, and so on indefinitely. 

In the expression for the length of the piece of the line 
left over, the first figure after the decimal point expresses the 
greatest whole number of tenths contained in that part; the 
next figure the greatest whole number of hundredths contained 
in the part (less than a tenth) which still remains, and so on. 

A number which cannot be exactly expressed by an in- 
teger or a fraction is called irrational. 

4. Positive and Negative Measures, It is a general rule 
of linear measurement, when expressed algebraically, that if 
a distance from a point in one direction is taken as positive, 
a distance from the same point in the opposite direction is 
regarded as negative. 

Thus if we represent the posi- I I I I I 

tive distance OA by the symbol x, ^ ^ o a b 
and if N, M, 0, A and B are equidistant points on a straight 
line, then 

OB = 2x; 

OM = ~ X', 

0N= - 2x. 

5. A single algebraic symbol, x for example, without sign, 
may represent either a positive or negative measure. 

If X represent a positive measure, then — x will represent 
an equal measure in the opposite or negative direction. 

If X represent a negative measure, then — x wiU represent 
an equal positive one. 




MEASUREMENT OF LINES AND ANGLES. 



Angular Measure. 

6. Def. An angle is the figure formed by two straight 
lines emanating from the same point, called the vertex of 
tlie angle. 

Def. The lines which form an angle are called its sides. 

7. Measures of Angles. An angle is measured by the 
length of a circular arc having its 
centre at the vertex of the angle and 
its ends on the sides of the angle. 

If the angle to be measured is / 
A OB, we conceive that with an arbi- / 
trary radius Oa an arc is drawn from { 
a to b. \ 

We regard as the positive direction \ 
that in which the arc is described by 
a motion opposite to that of the hands of a watch, and as the 
negative direction that in which the hands move. 

Hence we may consider the angle as measured either by 
the arc ab considered as positive, or the conjugate arc aMb 
considered as negative. The numerical sum of these two 
arcs is equal to a circumference. 

8. Choice of Unit. The absolute length of the arc will 
depend not only upon the magnitude of the angle, but upon 
the radius with which the arc is drawn. To avoid ambiguity 
from this cause, the unit of arc is supposed to be some fixed 
fraction of the circumference, and therefore greater the 
greater the radius. The arc is then indicated by the num- 
ber of units and parts of a unit which it contains, and this 
number is the same for the same angle whatever the radius 
may be. 

To indicate the angle corresponding to any arc we call it 
the angle of the arc, or, for brevity, the angle-arc. 

9. TJie Sexagesimal Division. The following is the usual 
division: 

The circumference is divided into 360 units, called degrees; 
Each degree is divided into 60 minutes; 
Each minute is divided into 60 seconds. 



4 PLANE TBIQONOMETBT. 

Then 

1 circumference = 360° = 21 600' = 1 296 000"; 

1 quadrant or right angle = 90°. 
This is called the sexagesimal division. 

10. Decimals of a Degree or Minute. Sometimes, instead 
of seconds, decimals of a minute are used. Both minutes and 
seconds may be dispensed with and decimals of a degree be 
used in their place. 

1 1 . Initial and Terminal Sides, In order to give entire 
algebraic precision to the measure of an angle, we must sup- 
pose a distinction between the side from which we measure 
and the side to which we measure. 

The side from which we measure is called the initial side. 

The side to which we measure is called the terminal side. 

We may conceive the initial side 
OA of the angle to turn round on 
until it reaches the terminal position 
OB. In thus turning, a point a upon / 
it will describe the circular arc which / 
measures the angle A OB. The length \ 
of the arc will then be proportional \ ^ 

to the amount by which OA turns \^ y 

in passing from OA to OB. ^^ "" 

The arc will be positive or negative according as the mo- 
tion is in the positive or negative direction. 

In the preceding examples we have supposed the initial 
side to be OA and the terminal side to be OB. Had we 
measured from OB to OA, the arc at would have been de- 
scribed in the negative direction, or aMb would have been 
described in the positive direction. Hence we should have had 

Angle BOA = — arc ah or + arc aMh, 
which is the negative of the corresponding measure from OA 
to OB. Hence 

By interchanging the initial and terminal sides lue change 
the algelraic sign of the angle. 

To give uniformity to this mode of measurement, the ini- 
tial side OA is supposed fixed, while the terminal side varies 
in direction according to the magnitude of the angle. 




MEASUREMENT OF LINES AND ANGLES. 



12, When angles are represented in a general way, the 
initial side may be conceived as extending out horizontally 
towards the right. The terminal side OB will then have a 
definite direction for each angle we choose to take. For ex- 
ample : 

For 90° the side OB will point upward. 
.. 180° '' '' ^•' '' '' to the left. 
cc 270° " " '' " '' downward. 
'' 360° '' " '' " '' to the right. 

13. Negative Measures. The initial side OA may reach 
the position OB by turning in the negative as well as in the 
positive direction. The measuring arc will then be (360° 
— arc al), and the negative measure will therefore be ex- 
pressed by 

- (360° - arc ah). 
Examples. 1. In the figure the angle AOB is 105°. 
Hence we may write 

Angle AOB = -^ 105° or - 255°. 
2. An angle of + 175° is the same as one of — 185°. 

a a a i 270° *^ ^^ ^' *^ *^ '^ 90° 

a (( a I Q50° " '^ '• '^ ^^ ^^ 10° 

etc. etc. etc. 

Counting the angles in the negative direction, 
For — 90° the side OB will point downward. 



180^ 

270^ 



to the left, 
upward. 



etc. etc. etc. 
In either case, as the angle ranges 
from zero to 360°, a point on the ter- 
minal side will move all the way round 
the circle. 

14. Division into Quadrants. The 
circle which measures angles is, for con- 
venience, supposed to be divided into 
quadrants, as in the figure. 

An angle between 0° and 90° is in the first quadrant. 
" '' '' 90° " 180° '^ '' second '' 

'' " " 180° " 270° '' '' third " 

" " " 270° " 360° ^•' '' fourth '' 




The four quadrants. 



6 PLANE TRIGONOMETRY. 

Counting the angles negatively from OA, 
An angle between 0° and — 90° is in iliQ fourth quadrant. 
" " '' - 90° '' - 180° '' '' third 

'' '' '' - 180° " - 270° " '' second 

" '' '' —270° '' —360° '' ''.first 

15. General Measure of an Angle, We may suppose the 
side OA to reach OB by moving positively through any 
number of revolutions plus arc al, or by moving negatively 
through any number of revolutions minus arc al. Hence, if 
i be the number of revolutions it makes, we may suppose 

Meas. of angle AOB = i X 360° + arc ah, 
or 

Meas. of angle AOB = — {i x 360° — arc ai) 
= — i X 360° + arc ab. 
Hence 

If we add or subtract any multiple o/360° to or from the 
measure of an angle, we leave the positions of its sides un- 
changed, 

EXERCISES. 

1. Erom the point emanate a set of 5 lines making 
equal angles with each other, and ^1. b fi> 
another set of 6 lines making equal \" / / 
angles with each other, the line OA ^\ ^\^ // 

being common to the two sets. Com- ^A /'' 

pute the values in degrees of the ten d~~ ~yv^ ""-^ 

angles A Ob, b OB, B Oc, etc., to/0.4. y"/ \ 

2. What is the value of that angle D / \\ 
whose negative measure is numeri- ^ \ \f 
cally double its positive measure? 

3. If a side starting from the zero point move through 
— 1905°, in what quadrant will it be found, and what will 
be the smallest positive measure of the angle? 

4. Two arms start together from the same position OA to 
turn round 0, the one going in the positive direction, so as 
to revolve in 60 seconds, the other in the negative direction, 
so as to revolve in 36 seconds. At what angle and in what 
time will they meet? 

5. If two revolving arms start out together from the posi- 
tion 0° in the same direction, the one going 5° a minute and 



MEASUREMENT OF LINES AND ANGLES. 7 

the other 8° a minute, through what arc will each have moyed 
when they again come together? At what angle will they 
meet? If they continue turning, after how many revolutions 
of each will they be together at their starting point? 

6. Four lines. A, B, O, D, emanate from the same point 
0, making angle B0C=2A0B, COD = 2B0C, DOA = 
2 COD. What are the values of the four angles which they form ? 

7. If an angle of 140° is multiplied successively by — 2, 
— 3, — 4, — 5, — 6, — 7, in what quadrants will the respec- 
tive multiples fall, and what will be the smallest positive 
measures of the several angles formed ? 

8. Show that the following pairs of angles are supple- 
mentary: 

90° + X and 90° - x-, ) 
270° - X and 270° + a;; V (Geom., §60.) 
60° - X and 120° -{- x. ) 

9. Show that the following pairs are complementary: 

45° - X and 45° + x; ) 
225° - x and 225° -}- x; l (Geom., § 59.) 
60° - X and 30° + x. ) 

16. Natural Measure of Angles. The division of the cir- 
cumference into 360° is entirely arbitrary, 
and any other angle than the degree may be 
taken as the unit. 

In purely mathematical investigations, 
where no division into degrees is required, 
the length of the radius is taken as the 
unit of measure. ^ ^°^£ ?S '"''''^ 

This unit is called the radian. ^^^ ^^ = ^^^"^ ^^• 

The radian is therefore the angle subtended by an arc 
whose length is equal to the radius. 

To find the relation of this unit of angle to the degree, 
minute and second, we note that the ratio of the entire cir- 
cumference to the diameter is 3.141 592 Q>o, etc. (Geom., 
BookVL, §484.) 

Hence its ratio to the radius is double this number, or 
6.283 185 3, etc. Since the circumference measures 360°, the 

unit radius will measure ^ ^oo -lo- o ;> or 57°. 295 779 5 . . . , 

b./doo loo o 




PLANE TRIGONOMETBT. 



Hence 


1 radian 


= 57.295 779 5 . . 


. . degrees. 






= 3437.746 77 . . 


. . minutes. 






= 206 264.806 . . 


. . seconds. 


In mathematics 


we use the symbol 




circumference 
"" diameter 


I circumference 
radius 


= 3.141592 65. 


Hence, 


, when we 


take the radian as 


the unit. 


7t 


represents 


an angle of 90°; 

U .f <C 13QO. 




'^Tt 


a 


" " '' 360° = 


: circumference; 


2n7r 


(( 


u c( a ^ circumferences. 






EXERCISES.* 





Considering the radius of the circle as unity, what is the 
length of circular arcs subtending the following angles? 

1. 28° 17' 15". 6. Ans. 0.4937. 

2. 14° 8'37".8. '' 0.2468. 

3. 22°'25' 53".4. '' 0.3915. 

4. 90°. " 1.5708. 

Note. In these exercises the angle is first to be reduced to a com 
mon denomination of measure, either degrees, minutes or seconds. 
For instance, 

28° 17' 15".6 = 101 835".6 = 169r.26 = 38°.287 67. 

If the radius is 100 metres, how many degrees and min- 
utes will arcs of the following lengths subtend? 

5. 100 metres. Ans. 57° 17' 44". 8. 

6. 72 metres. '' 41° 15'. 

7. 310 metres. '' 177° 37'. 

8. If an arc 32 metres in length subtends an angle of 32°, 
what is the radius? 

9. With what radius will an arc of the above length sub- 
tend an angle of 32'?. Of 32"? 

10. The smallest round disc clearly visible to the ordinary 
eye subtends an angle of 1'. What is then the greatest dis- 
tance at which a coin half an inch in diameter can be clearly 
seen? 

* The student should solve these exercises by the table of logarithms 
of numbers, the use of which should now be learned. 



MEASUREMENT OF LINES AND ANGLES. 9 

11. If a ball is to be visible at the distance of a mile, what 
is the inferior limit of its diameter? 

12. At what distance should a coin one inch in diameter 
be seen through a telescope magnifying 100 times? 

13. The sun being 93,000,000 miles away, what is the 
diameter of the smallest spot on its surface yisible to the 
naked eye? 

14 The moon being 240,000 miles away, what is the 
diameter of the smallest object on its surface visible through 
a telescope magnifying 600 times? 

15. The earth's equatorial diameter being 7912 miles, 
what is the length in miles of 1° of arc on the equator? 

16. The sun, at the distance of 93,000,000 miles, subtends 
an angle of 32'. What is its diameter ? 

17. The sun is 390 times as far from us as the moon. 
What is the greatest angular distance between the earth and 
moon as seen by an observer in the sun ? 

18. The satellites of Mars are about 8 miles in diameter, 
and were discovered at a distance of 37,000,000 miles. What 
would be the diameter of a ball in Boston to appear of the 
same diameter when seen from New York, distant 200 miles, 
that the satellites appeared when discovered ? 

19. The wheels of a bicycle are respectively 5 feet, and 18 
inches in diameter. Through what angle will the large wheel 
turn while the small one is making one revolution? ' ^ ~ 

20. A drunkard, instead of going in a straight line, de- 
scribes a series of semicircles whose centres and ends are all on 
the straight line in which he ought to walk. In what ratio 
is the length of his walk increased? 

21. Six equal circles, each of whose radii is r, touch each 
other and have their centres on the circumference of a seventh 
circle. What is the shortest distance by which one can walk 
round on the inner arcs of the six circles? 

22. If three circles, equal or unequal, mutually touch each 
other externally at the points A, B and O, to what constant 
is the sum of the angles subtended by the arcs AB, BO and 
CA equal? 



CHAPTER II. 

THE TRIGONOMETRIC FUNCTIONS. 



m. Line-functions of Arcs. Let AMX be a circle of 
any radius; OX and OM, any two radii, being the centre; 
XQ, the tangent to the circle at X, meeting OM produced 
in iV; MP, the perpendicular from M on OX. Then 
The line MP is called the sine of the arc MX, 
,c .. XN " " tangent '' 
'' '' ON " '' secant " *' 

Expressed in words, the definitions will be: 

The sine of an 
arc is the perpen- 
dicular from one ex- 
tremity of the arc 
upon the radius to 
the other extremity. 

The tangent of 
an arc is that por- 
tion of the tangent 
at one extremity of 
the arc which is 
contained between 
the point of tan- 
gency and the ra- 
dius through the 
other extremity. 

The secant of an arc is the line passing from the centre 
through the terminal point of the arc and terminated by the 
tangent at the initial point. 

The sine, tangent and secant are called line-functions of 
the arcs to which they belong. 




THE TRIGONOMETRIC FUNCTIONS, 



11 



EXERCISES. 

Prove the following theorems: 

1. The sine of an arc is haK the chord of twice the arc. 

To do this suppose the sine MP produced till it meets the circle 
below P, thus forming a chord. 

2. The sine of an arc measuring 30° is half the radius of 
the circle. 

3. The sine of an arc of 90° is equal to the radius. 

4. The sine has the same ratio to the tangent that the 
radius has to the secant. 

5. If the radius OM revolves around so as to become 
perpendicular to OX, the tangent and secant will each in- 
crease without limit. 

18, Trigonometric Functions of mi Angle. Using the 
same construction as before, let XOM be any angle whatever, 
and XM its measuring arc drawn with the arbitrary radius 
OX, Then 

The respective ratios of the sine, tangent and secant of 
the arc XM to the, radius OX are called the sine, tangent 
and secant of the angle XOM. 

19. By § 2 the ratio of either of these lines to the radius 
is a number expressing the length of the line on a scale of 
which the radius is unity. 

We have now to prove that 
this number is the same for the 
same angle whatever be the 
radius. 

Let X'OiV^' be the angle. 

Around the centre draw 
the two arcs XM and X'M' with 
any two radii OX and OX'. 

Erect the respective sines 
and tangents PM, XN, P'M, X'N'. 

Then because the triangles OPM, OXN, OP'M' and 
OX'N' have the angle at common, and the respective 
angles at P, X, P' and X' all right angles, and therefore 
equal, these triangles are equiangular and similar.. 




12 PLANE TBIGONOMETBT. 

Comparing the sides about the equal angles, we have the 
proportions 

PM : OM = P'M' : Oif' ; ) 

XN : 0X= X'N' : OX; [ (a) 

ON : 0X= ON' : OX. ) 

Because OM = OX = radius of inner circle, 

and 0M'= 0X'= radius of outer circle, 

we have, by definition, 

PM : OM = sine of POM; 

XN : OX = tangent of POM; 

ON : OX = secant of POM. 
The equations (a) now show that the sine, tangent and 
secant of the angle will be represented by the same numbers 
whether we measure them by the inner or outer circle. 
Therefore : 

I. To each angle of a definite magnitude corresponds 
One definite numher, called the sine of the angle; 
Another definite number, called the tangent of the angle; 
Another definite number, called the secant of the angle. 

II. These numbers may be represented by the lengths of the 
corresponding linef unctions on a scale of which the radius of 
the circle is chosen as the unit of lejigth. 

30. Notation. If we call a any angle. 

Its sine is written sin ^; 

^' tangent " tan a; 

'' secant '' , sec «. 

EXERCISE. 

Let the student find by actual measurement with diyiders 
and scale the sine, tangent and secant 
of every 10° from 0° to 90° in the fol- 
lowing way: 

With a radius equal to some unit or 
some whole number of units on a scale, ^ 
describe the quadrant XB. 

Divide the quadrant into 9 arcs of 
10° each. Through each point of divi- 
sion, MioY instance, draw a radius and 
continue it until it intersects the tan- 
gent at N. Then measure — 




THE TRIGONOMETRIC FUNCTIONS, 



13 



1. The distance of each division point on the arc from the 

line OX, which distance divided by the radius will give the 
sine of the corresponding angle. 

2. The distance of each point of intersection, N, from X, 
which being divided by the radius will give the tangent of the 
angle. 

3. The length of each ON, which being divided by the 
radius will give the secant of the corresponding angle. 

The results should all be expressed in decimals to three 
places, exhibited in a little table in the following form, and 
afterward compared with the values found in the trigonome- 
tric tables: 



Angle. 


Sine. 


Tangent. 


Secant. 


0° 








10° 


. . . 


. . . 


. . . 


20° 


. . . 


. . . 


. . . 


30° 




. . . 


. . . 


etc. 




. . . 


. . . 



With care the average deviation of the measures from the 
truth ought not to exceed .005, -except in the cases of the 
tangent and secant of 70° and 80°, which are so great that 
they cannot be easily found in this way. 

21, The Cosine, Cotangent and Cosecant. In the pre- 
ceding sections we have supposed the initial side of the angle 
to go out toward the right, and the positive direction of 
motion to be opposite to that of the hands of a watch. But 
this restriction is only to fix the thought. We may suppose 
the angle to have any situation and to be counted in either 
direction without changing the values of the sine, tangent 
and secant, provided that we take the positive directions of 
the lines representing the functions to correspond to the 
direction in which the angle is reckoned. 

Let us count the angle from OJTin the direction toward 
OX. The tangent line must then touch the circle at Y, and 
its positive direction must be toward the. right. 



14 



PLANE TBIGONOMETBT. 




Then the lines 

PM= sin POM, \ 

YN = tan POM, [ (radius OF = 1) (b) 

0N= sec POM,) 
will have the same values as in an angle equal to POM 
counted in the usual way from OX toward Y. 

Now because angle XOM + 
angle MOY =^ angle XOY = 90°, 
MO Y is the complement of XOM. 
Therefore the equations {h) may be 
written 

PM represents sin comp. of XOM-, 
YN represents tan comp. of XOM-, 
ON represents sec comp. of XOM. 
The sine, tangent and secant of 
the complement of an angle are regarded as three additional 
trigonometric functions of the angle itself. They are named 
thus: 

The sine of the complement is called the cosine of the angler 
The tangent of the complement is called the cotangent 
of the angle. 

The secant of the complement is called the cosecant of 
the angle. 

Thus the new functions are defined in the following way: 
cosine a = sin (90° — a);^ 
cotangent a — tan (90° — a)\y (1) 

cosecant a = sec (90° — a). ) 
The words cosine, cotangent and cosecant may be ab- 
breviated to cos, cot, CSC, respectively. 

23. The forms (1) enable us to find the cosine, co- 
tangent and cosecant of an angle when we know the sine, 
tangent and secant of its complement. Thus if the cosine of 
60° is required, we have 

cos 60° = sin (90° - 60°) = sin 30°. 
Also, by substituting 90° — a for a, we find 
cos (90° — a); 



sm a 



tan a = cot (90^ 
sec a = CSC (90' 



a); 
a). 



(3) 



THE TRIGONOMETRIC FUNCTIONS. 



15 



23. The Versed-sine and Co-versed-sine. Besides these 
six functions, two others, the versed-sine and co-versed-sine, 
are sometimes used. Their definitions are: 

Versed-sine = PX = 1 — cosine; 

Co-versed-sine = F — PM = 1 — sine. 

24, The six trigonometrical functions may be represented 
on a single dia- 
gram. The func- ^ ^^^^,^^^^^ ^,^ 
tions named in the 
diagram are all 
those of the angle 
XOM. For the se- 
cant and cosecant 
we have 

ON = sec XOM, 
ON = CSC XOM, 
because XOM is 
the complement of 
*MOY, the secant of 
which is 0N\ 

Because PM\\ OP' 

and OP II P'M, 

therefore OP = P'M, 

so that we may take either OP or P'M to represent the cosine 
of XOM. 




EXERCISES. 



1. Of what angle is the cosine equal to the sine of 85° ^ 



ii 35° ? 
cosine of 25° ? 



2. " " '' '' cosine '' 

3. " " '' '' sine '' 

4. " " " ^^ secant '' '' cosecant of 22° ? 

5. Prove that cos (45° + ^) = sin (45° - B)-, 

cot (45° + ^) = tan (45° - B); 
CSC (45° + 5) = sec (45° - B). 

6. Of what angle is the sine equal to the cosine? 

7. Between what limits must an angle be contained when 
its cotangent is greater than its tangent, and vice versa? 

8. Is it possible for both the tangent and cotangent of an 
angle to be less than 1 ? 



16 



PLANE TBIQONOMETBT. 




Relations between the Six Trigonometric 
Functions. 

25. When we know any one trigonometric function of an 
angle a yalue of the angle can be determined. Knowing the 
angle, the values of 

the other functions -y ^^^^^^.^ j^< 

can be found. Hence 
if any one function of 
an angle is given all 
the others can be 
found. We have now 
to investigate the al- 
gebraic relations by 
which this may be 
done, seeking to ex- 
press each function 
in terms of the five 
others. 

Let us put 

a = angle XOM, 

Because OF Mis a right-angled triangle, 
OP' + PM' =:z 0M\ 

\OMJ ^ \0M) ' 
cos'^ a -\- sin^ a = 1^ 
cos a =± V 1 



or 



Hence 
which gives 
and 



sm a 



(3) 
(4) 



sin «: — ± r 1 — cos'^ a. 
•the similar triangles 0PM and OXN give 
OP :PM= OX'. XN', 
OP: 0M= OX: ON. 
Substituting for the lines their algebraic equivalents, the 
first proportion gives 

cos a : sin « = 1 : tan a. 

sin a sin a ,^. 

(5) 



Hence 



tan a = 



cos Oi 4/1 _ siii2 ^ 

which gives the tangent in terms of the sine and cosine. 
The second proportion gives 

cos a: : 1 = 1 : sec Of, 



THE TBIOONOMETBIC FUNCTIONS. 17 



whence sec a = =-— = . (6) 

cos a: |/i _ sm^ a ' 

We conclude from this : 

Tlie product of the cosine and secant is equal to unity. 

In other words, the secant and cosine are reciprocals of 
each other. 

By a similar course of reasoning upon the complementary- 
triangle we find that the cosecant and sine are reciprocals of 
each other. 

The similar triangles OP 'if and OYN' give 
OP' :P'M= OY: TJV'; 
but OP' = FM= sin a (when 0F= 1), 

whence sin a : cos a = 1 : cot a, 

, cos a Vl — sin^ a fn\ 

and cot a = = -. (7) 

sm a sm a 

Comparing with (5), we have the relation: The product of 
the tangent and cotangent of any angle is unity. 

In other words, the tangent and cotangent are reciprocals 
of each other. 

26. We thus reach the general conclusion that the three 
complementary functions are each the reciprocal of one of the 
three other functions, namely: 



cosme = --; 

secant 

cotangent = 



tangent' 
cosecant = 



sine 



(8) 



EXERCISES. 

1. If sin y = 0.60, find cos y, tan y, cot y, sec y and 
cosec y. 

2. Find the values of the same functions when cos y = 
0.60. 

3. Prove that the mean proportional between a cos x and 
a sec X is a. 

4. Prove that the mean proportional between a tan /3 and 
b cot p is Vab, 



18 



PLANE TBIOONOMETBT. 



5. Express sin 25° in terms of sec 65°. 

6. Prove tan (45° - /?) X tan (45° +y5) = 1. 

37. Special Values of Trigonometric Functions. If angle 
XOM= 45°, we shall also have OMP = 45°, and therefore 

OP' + PM' = 2PM' = 0M\ 
whence PM=. OM Vi. 

Therefore 

PM : OJf = sin XOM= sin 45° = vT- («) 
In the same way we find 
XT = OX, 
whence 
XT : 0X= tan XOM=: tan 45° = 1. (i) 

V20X; 




Again, 
whence 



0T=V0X'-\-XT' 



sec 



XOif=sec45°:= /2. 



(^) 



The values of the complementary functions are next found 
by the equations (8). We then have the following values of 
the six functions of 45°: 

sin 45° = cos 45° = ^. 

tan 45°= cot 45° = 1. 

sec 45° = cosec 45° = j/^. 

28. Functions of 30°. Next, let angle XOM = 30°. 
Make angle XOM' = XOM = 30°. 
The triangles MOM' and TOT' then 
have each of their angles 60°, and so are 
equilateral. Therefore 

MP = WM' = iOM 
Hence 

PM: Oilf=sin30° = i. 

In the same way, supposing OX equal to unity, 

XT=iOT^.* 
OT' - XT' = OX' = 1; 

iOT' = l, ^"^ ■'"' ^ 



iOX. 




or 



0T= Vi = 



vw 



XT 



1 

V3. 



Also, 



OP = VOP' - MP' = V1- i^ 



V3 



, 



TEE TRIGONOMETRIC FUNCTIONS. 19 



Hence sin 30° = -^, 



tan 30° 



1 

vr 



sec 30° = ~, 

V3 

cosec 30° = 2, 
cot 30° = -^3; 

id 

cos 30° = -^-. 

29. Functions of 18°. It is shown in geometry (§§ 464, 
465) that if the radius of a circle be divided in extreme and 
mean ratio, the greater segment will be the chord of 36°; 
that is, twice the sine of the arc of 18°. 

Putting 1 for the radius and r for the greater segment, 
the condition that the division shall be in extreme and mean 
ratio is 

1 : r : : r : 1 — r, (Geom., § 440) 

or, equating the product of the means to that of the extremes, 
r^ = 1 — r. 
The solution of this quadratic equation gives 
- 1 ± 1/5 

/ = ^ ■ 

The positive root is the only one we want. Hence, because 
r is twice the sine of 18°, 

• -too V5-1 
tr = sm 18 = . 

We then find cos'^ 18° = 1 - sin^ 18°. 



XT , oo '^10 + 2 4/5 
Hence cos 18° = ^ — - — 



Note. At this poi^t in his course the student should learn the 
nature and use of the Tables of Natural and Logarithmic Trigonometric 
Functions, as explained at the end of the present book, or in the 
5-place tables. 



20 PLANE TBIOONOMETBY. 

EXERCISES. 

Express the following functions as functions of the com- 
plementary angles: 

1. cos 55°. Ans. sin 35°. * . ' 10. cos (45° + x). 

2. sin 55°. 6. sec 5°. , -. 11. tan (90° - x), 

3. sin 45°. 7. esc 12°. . 12. tan (25° + x). 

4. sin 85°. 8. cot 23°. 13. cos (25° - x). 

5. tan 25°. 9. sin (45° + x). 14. sec (65° + x). 

15. By means of the preceding formulae, §§ 27-29, com- 
pute the yalues of the sine, cosine, tangent and cotangent of 
45°, 30° and 18°, and see if the numbers and logarithms given 
in the table are correct. 

The computation for 18° is made as follows, in part: 
log 5, 0.6990 SVs; 4.472 log (10 + 2^5), 0.5803 

log 4/5; 0.3495 add 10, 14.472 log 4, 0.6021 

log 2, 0.3010 log, 1.1606 log cos 18°, 9.9782 

log 21/5; 0.6505 log 4^10 + 3 4/5; 0.5803 cos 18°, 0.951 

16. If the cosine of an angle is m times the sine, what are 
the values of the tangent and cotangent? 

17. Of what angle is the sum of the sine and cosine 1.25? 
(We substitute for either the sine or cosine its value (4) in 

§ 25. Thus, if we put s for the sine, we have the equation 

s + VV^^s' =. 1.25, 
which, being solved, gives the value of s from which the 
angle is found by the tables.) 

18. If the sum of the sine and cosine of an angle is n, ex- 
press the separate values of the sine, cosine and tangent. 

19. Of what angle does the sum of the cosine and secant 
make 3? (§26.) 

20. Erom the equations (8), (5) and (3) prove the rela- 
tion 

sec a = Vl -\- tsui^a. 

21. The sum of the tangent and secant of an angle is n. 
Express their separate values in terms of n. 

Ans. tan = --r , and sec = -—: . 

2n 2n 



1 



THE TEIGONOMETBIG FUNCTIONS. 21 

22. Of what angle does the cosine exceed the sine by 0.5? 

23. The sum of the tangent and cotangent of an angle is 
m. Find the separate yalue of the tangent and cotangent. 

24. Find an angle of which the sine is twice the cosine. 

25. Of what angle is the tangent double the sine? 
Keduce the following expressions to their simplest form: 

26. tan (40° + a) tan (50° - a). 

27. tan (45° + a) cot (45° - a), 

28. cos (45° + a) sec (45° - a), 

29. tan a cos a. 

- 30. *5ILf. 31. 5^*. 

cot X sec X 

' 32. (sin X -\- cos xY + (sin x — cos xy. 

- 33. {a smx-\- b cos x) (d sin x -{- a cos x) 

-\-(asmx — d cos x) (b sinx — a cos x). 

- 34. sin (20°+«:) cos (70°-«) + cos (20°+a) sin (70°-«r). 
' 35. siji'a Gos^ J3 + cos^a sin^/? + sin^a: sm^j3 -\- cos'^a cos^/?. 
" 36. (1 — sin a) (1 -f sin a) -\- (1 — cos a) (1 + cos a). 

- 37. (sec d + tan 0) (sec ^ - tan 0). 

38. Vtan^ a: + cot' x -{- 2 Vtan^ ^ + cot' x - 2. 

- 39. sin':z;(l + cot'a;). 

^ 40. sec CSC — tan 6 — cot 6. 

41. If sin ii; = m sin y and tan x = n tan ?/, prove 

,2 ^i' — m"* 
tan cc = — T z~. 

7)f — 1 

42. Proye the equation 

?^(cot'0-cos^6^) = l. 

. cos 6 ^ ' 

43. Prove that cot' B - cos' 6 = cot' cos' 0. 

44. When tan x -\- cot ic = 2, find sin x + cos a;. 

45. What is the square root of cos' x -{- 2 ~\- sec' x? 

46. From the results of § 28, write the values of the six 
functions of 60°. 

47. From the results of § 29, write the values of the six 
functions of 72°. 

48. Of what angle is the tangent double the cosine? 

49. Of what angle is the tangent half the cosine? 

50. Express sec a in terms of cot a. (Oomp. Ex. 20.) 



CHAPTER III. 
OF RIGHT TRIANGLES, 



30. Fundamental Relations. Let OCN be a right tri- 
angle of which, a and h are the sides which contain the right 
angle, C the hypothenuse, and a and /? the angles opposite a 
and h respectively. 





C X 



If we take ON as a radius and draw the arc NX from the 
centre 0, the side NC will, by definition, represent the sine 
of XON, and OG its cosine, when the radius is ON, That is. 



NG 
ON 



= sm a: 



OG 
ON 



cos Of. 



(1) 



We may show in the same way, by taking N as the centre 
and NO as the radius. 



OG 



= sin /3; 



NG 



= cos /?. 



(^) 



ON "^^ ON 

We might also have deduced these equations from (1), be 
cause /? = 90° — a, whence 

sin J3 = sin (90° — a) = cos a', 
cos /? = cos (90° — a) = sin a. 
Again, by taking OCas a radius, we find 



GN 
OG 



= tan a = cot /3; 



OG 

GN 



tan /? = cot or. (3) 



/ 



BIGHT TBIANGLHS. 



23 



(4) 



Putting NO = a, 00 = h, ON = c, the equations (1), 
(2) and (3) give the relations 

a=^ c sin a ^=1) tan a-, 
^ = c cos a = a cot a; 
c = a CSC a = b sec a. ) 
We may express the same relations in terms of /3, using 
the complementary functions as follows: 

a = c cos fi = b cot /3; \ 
d = c sin /3 =: a tan J3; V (5) 

c = a sec J3 = d CSC /3. ) 
31. These relations may be summed up in the following 
general theorems: 

I. The hyioothenuse of any right tria7igle is equal to a 
side into the secant of its adjacent angle or the cosecant of its 
opposite angle. 

II. A side is equal to the hypothenuse into the sine of the 
opposite angle or the cosine of the adjacent angle. 

III. One side is equal to the other side into the tangent of 
the angle adjacent to that other side or the cotangent of the 
tingle adjacent to itself 

EXERCISE. 

Show by the aboye equations how each side will be ex- 
pressed in terms of the others when a = 30° and when a = 45°, 
using the values of sin a, etc., already found (§§ 27, 28) — 
namely, 

sin 45° = Vi; tan 45° = 1; etc, 

and sin 30° = -J; cos 30° = -^; tan 30° = -^; etc.— 

and from the results prove Geom., §310, and show that a tri- 
angle each of whose angles is 60° is equilateral. 

33. Examples and Exercises in Exptression. 
companying figure OQN, ONP q^ 
and OXN are right angles, and 
we put 

a = angle XOW; 

/? = angle NOQ. 

It will also be noticed that 

Angle XNP = a 

and Angle OJVX = XPN= 



In the ac- 




24 ' PLANE TRIGONOMETRY. 

It is now required to express all the other lines in terms of 
OX and trigonometric functions of a and /?. 
Solution. We have 

0]Sf=: OX sec or; 

XN= OXid^na-, 

OP = OiV^sec a = OX sec' a; 

NP = OP, sin a — OX sin a sec'* a = OX ism a sec a; 
or iVP = OiVtan a = OX sec a tan a (as before); 

0§ = OX cos /? = OX sec a cos /3; 

NQ = OXsm /3 = OX sec asinfi; 

EXERCISES. 

1. By the same process express OQ, QX, OX, XX, NP 
and XP in terms of ON and trigonometric functions. 

2. Express the same quantities in terms of NP. 

3. Express XX separately in terms of OX and XP, and 
by multiplying the two values prove the theorem that XX is 
a mean proportional between OX and XP (Geom., § 401). 

4. In a right triangle the sides which contain the right 
angle are a and d, {ayh), and 6 is the difference of the angles 
at the base. Express the length of the perpendicular from 
the vertex upon the base in two ways, and the lengths of 
the segments into which it divides the base each in one way. 
The expressions are all to be in terms of a, h and d. 

Ans. (in part). One expression for the perpendicular is 
p = asm (45° -i^). 

Solution of Right Triangles. 

Since in a right triangle one angle — the right angle — ^is 
given, only two other independent parts are required to solve 
the triangle. These two parts may be any two of the sides or 
one side and one angle. What parts soever are given, the re- 
maining parts may be found by the equations (4) and (5). 
The following are all the essentially different cases. 

33. Case I. Given the two sides a and l adjacent to the 
right angle.'^ 

* It is recommended that in commencing this subject the student 
first solve a few of the problems by natural numbers, using either such 



BIGHT TRIANGLES. 25 

The first equation (4), 

a = b tan a, 

, a 

gives tan a = j. 

Therefore the quotient of the two sides gives the tangent 
of the angle opposite the dividend side. 

From the tangent the angle a is itself found by the trigo- 
nometric tables; then sec a or cos a ; then the hypothenuse c 
from the equation 

-L ^ 

cos a 
Example, Given « = 9 metres, J = 12 metres, to find 
the remaining parts of the triangle. 
Solution hy Natural Numhers. 

tan « =y93- = 0.75. 
Looking in the table of natural sines, etc., we find that 
the angle whose tangent is 0.750 is 36°.85. Hence a = 36°. 85. 
The third angle, /? = 90° - «f = 53°. 15. 
sec 36°.85 = 1.250. 
c = ^^ sec 36°.85 = 15. 

Logarithmic Solution. 
log a, 0.9542 log cos «, 9.9031 



log J, 1.0792 


logb, 1.0792 


logtanrt', 9.875 


logc, 1.1761 


a, 36° 52' 


c, 15 metres. 


ft, 53° 8' 




EXERCISES. 


1. Given a= 4, b = 


5; find remaining parts. 


2. '' a= 8, b = 


5; 


d. '' a= 43.148, b = 


84.107; " '' 


4. '' a= 2.7938, Z'^: 


876.59; '' 


5. '' a= 759.48, d = 


51.85; " 


6. '' a= 8628, b = 


27 316; 


The first two exercises are made 


purposely simple, that they may be 


performed without logarithms. 





tables as he may construct for himself by measurement as described in 
§ 20 or the three-place tables of natural sines, cosines, etc. 



26 PLANE TRIGONOMETMY, 

34, Case II. Given the Jiypotlienuse and one side, • 
Solution. From the equation 

c mi a = a 

we obtain sin a = —, 

c 

which may be used to find a when a and c are given. Then 

the remaining side is found by the equation 

h = c cos a. 

Example. Given a — IZ, c = 20, to find the remaining 
parts. 

Solution ly Natural Nunibers. 

sm Of = — = TT-r = 0.65. 
c 20 

From the table we find a = 40°. 5. 



Logarithmic Solution 

log 13, 1.1139 
log 20, 1.3010 

log sin o', 9.8129 
- a, 40° 33' 

A 49° 27' 



EXERCISES. 

1. Given « = 7, c = 9; find remaining parts. 

2. "5=9, c = 16; 

3. " « = 82.143, c = 120.412; " 

4. " J= 2.9235, c= 9.827; '' 

5. A circle of radius r is drawn with its centre at a dis- 
tance p from a straight line. What length will it cut out 
from the line? What will be the result if r < p? 

35. Case III. Given an angle and one side, as a and a. 
Solution. The equations (a) give 

h = =z a cot a = a tan 6; 

tan a 

a a 

a CSC a = a sec p. 



cos flf, 

log 20, 


9.8807 
1.3010, 


log J, 


1.1817 
15.20 



sm a 



BIOHT TRIANGLES. 



27 



Note. One oblique angle being given, the remaining one may be 
found from the equation 

yS = 90° - a. 



EXERCISES. 



39^ 



19.5: 



10.925: 



find a and h. 



8.1273; find « and (?. 
0.9271; '' '' 



11' 



-P 



\\\ 



What are the expressions for the distances AP and 
will be the distance if -4 (7 = 80 yards and a — 



Giyen a = 72' 6\i' , c 

" a= 16° 25', c = 

" /? z= 43° 28', b = 

" /3 = 8° 29'.2, d = 
An engineer, desiring 
to find the distance from a 
point A on one bank of a riv- Af- 
er to a point P on the other 
bank, measured off a base 
line AO, h yards in length, 
in a direction perpendicular 
to AP. He then measured 
the angle A CP, and found it 
to be a. 

cp-^. 

What 
85°.22'.5? 

6. An engineer, desiring to determine the height of a yer 
ticalwall, measured off a 
distance PO = 1) feet on 
leyel ground, and then 
from a point E, a feet 
aboye the ground, mea- 
sured the angle HEX = 
a between the line of 
sight EX and the hori- 
zontal line EH. Ex- 
press the height PX of "^ ^' 
the wall algebraically in terms of a, b and a, and compute 
the height when h = 400 feet, a = 6 feet and the angle a = 
22° 17'. 

Method of Solution. Find the height HX and add HP 
= a. 



/ 



H 






28 PLANE TBIGONOMETBT. 

7. Desiring to find the height of an inaccessible rock 
M aboYe a plane, 
its angle BPM wsis 
measured from a 
point F and found 
to be a. The ob- 
server then advanc- 
ed a metres towards _ 
the rock to §, and P « Q ^ B 

there again measured the angle of elevation and found it to 
be y. Express the height BM of the rock and compute it 
when a = 2000 metres, a = 30° 28' and y = 40° 53'. 

*" Method of Solution, Let the vertical height BM = h and QB = x. 
Then we have the two right triangles PSif and QBM, which give 
h = {a -\- x) tan a ; 
h ■= X tan y. 
From these two equations we obtain the following values of h and x: 

a tan a 




h = 



tan y — tan a 
a tan a tan y 
tan y — tan a 



8. The altitude of a triangle is 7.2648, and the angles at 
the base are 72° 29.3' and 40° 30. 5' respectively. Compute the 
base and sides. Also, find the general expression for the 
length of the base in terms of altitude and angles at base. 

9. From the top of a tower 108 feet high the angles of de- 
pression of the top and bottom of another tower standing on 
the same horizontal plane are found to be 28° 56' and 53° 41' 
respectively. Find the distance between the towers, the 
height of the second tower, and the distance between the 
summits of the two towers. 

10. In a circle of radius r, express the length of each side, 
and of the apothem, of a regular inscribed polygon of n sides. 
Find first the special values for the triangle, square, pentagon, 
hexagon and octagon. 

Ans. For the octagon: side = 2r sin 22-J°; 
apothem = r cos 22^°. 

11. If the side of a regular octagon is 10 metres, what are 
the radii of the inscribed and circumscribed circles? 



BIGHT TRIANGLES, 29 

12. At what altitude is the sun when a tower 20 metres 
high casts a shadow 75 metres long upon a horizontal 
plane? 

13. A house 30 feet square has a roof with a slope of 30° 
which projects 6 inches horizontally oyer the sides of the 
house on all sides. How many square feet of tin will be re- 
quired to coyer the roof ? 

14. It was found that the length of the shadow of a 
monument upon a horizontal plane diminished 22 metres 
Avhen the sun's altitude increased from 30° to 45°. What was 
the height of the monument ? 

15. If the sides of a rectangle are represented by the num- 
bers 5 and 1, what will be the length of the diagonals, and 
what angle will they form with each other? 

16. If /? is one of the acute angles of a right triangle, and 
c its hypothenuse, express the altitude in terms of c and fi. 

17. The altitudes of two lighthouses, of equal height and 500 
metres apart, are measured by a ship in line with them, and 
found to be 12° 20' and 7° 30' respectiyely. Find their com- 
mon height aboye the ship's deck, and the distance of the nearer. 

18. In the figure of § 32 suppose 

Angle a = 36° 24', 
Angle /3 = 41° 58°, 
LineiVX= 1, 
and thence find the lengths of all the other lines in the figure. 

19. Find the side and the area of the square inscribed in 
the circle whose radius is unity. 

20. Find the side and the area of the regular pentagon in- 
scribed in the circle of radius unity. 

21. Find the side and the area of the regular pentagon 
circumscribed about the circle whose radius is r. 

22. The perpendicular from the right angle of a triangle 
^0 the hypothenuse diyides the latter into segments whose 
lengths are 9 and 16 respectiyely. Find the sides and angles 
of the triangle. 

23. A riyer one mile wide flows at the rate of 4 miles an 
hour. A steam ferry-boat with a speed of 8 miles an hour 
has to run directly across. In what direction must she steer, 
and how long will she be in crossing? 



30 PLANE TBIGONOMETBT. 

24. The equal sides of an isosceles triangle are eacli 
Vb -\-l, and each of the adjacent angles is 72°. Find the 

base and altitude. (§ 29.) 

25. If the earth be a sphere whose diameter is 7912 miles, 
what are the radii of the parallels of latitude 18°, 30°, 45°, 
60° and 72° respectively? 

26. The Washington monument is 555 feet high. If a 
man 800 feet east from the centre of the base walks 600 feet 
north, what will be his distance and direction from the base, 
his distance from the summit, and the angular elevation of 
the summit? 

27. In an isosceles right triangle the sum of the hypothe- 
nuse and one side is 3. Find the sides. 

28. The perimeter of a right triangle is 6, and one oblique 
angle is 30°. Find the three sides. 

29. The hypothenuse of a right triangle is 8 metres longer 
than the shorter side and 1 metre longer than the longer side. 
Find the three sides and the angles. 

30. The perpendicular from the vertex upon the hypothe- 
nuse of a right triangle divides the hypothenuse in the ratio 
1 : 3. Find the angles. 

31. The hypothenuse of a right triangle is 13 metres, and 
the altitude above the hypothenuse is 6 metres. Find the 
sieves and the angles. 

32. If "the diameter of a carriage-wheel is 5 feet, at what 
angle does a spoke stand whose outer end is 2 feet above the 
ground? 

33. The wheel of a bicycle is 54 inches diameter. A mark 
is made at the bottom of the wheel, and the bicycle is wheeled 
100 inches. How high is the mark then above the ground? 

34. If a wheel 8 feet in interior diameter has 10 spokes, 
what is the distance between the ends of the spokes? 

35. How far is a pitcher at base-ball from the first and 
third bases, if the latter are 90 feet and the pitcher 45 feet 
from the home-base? 

36. A man standing 120 feet north from a monument 100 
feet high wants to walk east until the elevation of the top of 
the monument from the ground where he stands is 30°. How 
far must he go? 



EIGHT TRIANGLES. 31 

37. A ship sights a light-house bearing due north. After 
sailing north 40° east, 15 miles, the light-house bears west. 
What was the distance of the light at each bearing? 

38. A man on top of a monument in aleyel plain finds the 
depression of a certain point on the plain to be 28°. Descend- 
ing 100 feet he finds the depression to be 15°. What is the 
height of the monument and the distance of the point from 
its base? 

39. If the radius of the earth is 3956 miles, what angle 
would the earth subtend if viewed from a point 6000 miles 
above its surface? 

40. If the diameter of the sun is 860,000 miles, that of the 
moon 2160 miles, and the distance of their centres 93,000,000 
miles, what is the length of the moon's shadow, measured 
from the centre of the moon to the apex of the shadow? 

41. A ladder 55 feet long will reach a window 28 feet high 
on one side of a street, or 37 feet high on the other side, its 
foot remaining in the same position. What is the breadth of 
the street? 

42. In a right triangle one of whose angles is 30°, the sum 
of the hypothenuse and altitude is 12.5. Find the three 



43. If a sheet of paper 42 X 30 inches be cut across diago- 
nally, what is the distance of each corner from the cut? 

44. A room 28 feet from IST. to S., 18 feet from E. to W., 
and 14 feet high, has a string tightly drawn from the lower 
N.E. corner to the upper S.W. corner. What is the shortest 
distance of the upper N.E. corner from the string? 

45. The hypothenuse of a right triangle is 3 times the al- 
titude. Find the angles. 

46. The hypothenuse of a right triangle is 25, and the 
sum of the other two sides is 31. Find the angles. 

47. In a circle the sum of the chords of 60° and 120° is 3. 
What is the radius of the circle? 

48. If the sides of a rectangle are 7 and 10, what angle will 
the diagonals form with each other? 

49. The length of the diagonal and longer side of a rect- 
angle is 18; that of the diagonal and shorter side is 16. Find 
the sides, and the angle at which the diagonals intersect. 



32 PLANE TRIGONOMETRY. 

36. Projections of Lines. 

Theoeem. The projection of a line is equal to its length 
multiplied hy the cosine of the angle which it forms with the 
line on which it is projected. 

Proof. Let the finite line ^^ be projected upon the in- 
definite line XY (Geom., § 280). 

By definition the projection 
will be the distance XY be- ^^ 

tween the feet of the perpendic- ^^ 

ulars from A and B. ^^ 

Through one end of the line, ^.f^^^- 

as A, draw A G parallel to XY, \ 
meeting J5 J^ in G. Then I 

J[(7=XF; ^ 

AG=ABgo^BAG. 
Hence Projection XY ^ AB cos BA G. Q. E. D. 

EXERCISES. 

1. If we put 

r = the length AB, 

c = the length XA, 

a = the angle GAB, 
it is required to express the lengths of the diagonals XB 
and A Y (not drawn in the figure), and the tangents of the 
angles which they make with XY in terms of r, c and a, 

2. If the ridgepole of a roof inclined 30° to the horizon 
is distant a feet from the eayes, what is the distance of two 
plumb-lines, one dropped from the end of the eayes, the other 
from the end of the ridgepole? 

3. The sine of one acute angle of a right-angled triangle is 
equal to the tangent of the other acute angle. What are the 
acute angles? 

Note. Use formulae (2) and (3), § 25. 

4. Of what angle is the cosine equal to the tangent? 



CHAPTER IV. 

TRIGONOMETRIC FUNCTIONS OF UNLIMITED ANGLES. 




37. "We haye hitherto considered only trigonometric 
functions of angles less than 90°. We have now to define 
these functions for angles of any magnitude. 

As in Chapter I. , let the initial side OX of the angle re- 
main fixed, while the terminal side 
Om revolyes around the centre 0, 
so that the point m passes succes- 
sively through Y, M, X', M' , Y', 
M" and X. "We now consider the 
values which each trigonometric 
function assumes as the terminal 
side goes through a complete revo- 
lution. 

In this study we shall assume 
that the radius Om is unity, and 
that all lengths are represented in terms of this unit, so that 
the trigonometric functions shall be represented by their ap- 
propriate lines. 

38. Changes in the Sine. The general definition of the 
sine is this: 

The sine of the angle is that number which expresses the 
length of the perpendicular from the end of the terminal line 
upon the initial line. 

In order that the perpendicular may always meet the ini- 
tial line, the latter must be supposed produced to X', making 
a complete diameter of the circle. 

When the perpendicular is below the initial line the num- 
ber expressing its length is algebraically negative. 

Following the movement of the point m around the circle, 
we have the following results: 



34 



PLANE TBIOONOMETBY. 



First Quadrant. As the angle increases from 0° to 90° 
the sine increases from to 1. Hence 
sin 90° = 1. 

Second Quadrant. As the angle increases from 90° to 
180° the sine diminishes from 1 to 0. Hence 
sin 180° = 0. 

Third Quadrant. As the point m moves from X' to Y, 
the perpendicular to XX' must be drawn upward. Hence 

In the third quadrant the sine is negative. 

At Z"' the sign is — 1. Hence 

sin 270° = -1. 

Fourth Quadrant. From Y' to X the sine changes from 
— 1 to 0. Hence 

In the fourth quadrant the sine is negative, 
and sin 360° = sin 0° = 0. 

The changes of algebraic sign as the 
angle goes through the four quadrants are 
shown in the annexed diagram. 

Angles having Equal Sines. If angle 
X'OM = XOm, the two angles XOm and 
XOM will be supplementary. Also in this 
case the triangles pOm and POM will be 
identically equal; so that PM = pm. 

Now PM represents by construction the sine of XOM, 
and pm the sine of XOm. Hence 

The sines of supplementary angles are equal. 

In symbolic language this theorem is expressed thus: If a 
be any angle, then 

sin (180° - «') = sin « • (2) 

and sin (90° -^ a) =. sin (90° — a). 

If the points M' and M'' are equally distant from Y', so 
that angle M' OY = angle Y'OM", which angle call y, the 
sines P'M' and P"M" will be equal. Hence, whatever be y, 
sin (270° - y)~ sin (270° + ^). 

These results may be summed up as follows: 

39. OoROLLAEY. To a given sine may correspond 
of two angles. 

The sum of these tioo angles is 180° or 540°. 




Algebraic signs of 
the sine. 



FUNCTIONS OF UNLIMITED ANGLES. 



35 




40. Changes in the Cosine. The general definition of 
this function is : 

The cosine is represented hy the distance of the foot of the 
sine from the vertex of the angle. 

The cosine is positive when the perpendicular falls upon 
the terminal side; negative when it falls on this side pro- 
duced. 

First Quadrant. \i Dm takes 
the zero position OX, p will fall 
upon X and Op will become equal 
to 1. Hence 

cos 0° = 1. 

As the angle increases from 0° x'[- 
to 90°, p moYCS from X to 0, and 
the cosine diminishes from 1 to 0. 
HenCe 

cos 90° = 0. 

Second Quadrant. As the 
angle increases from 90° to 180°, p moYes from to X', 
on the negative side of 0. Hence 

In the second quadrant the cosine is negative, and 
cos 180° = - 1. 

Third and Fourth Quadrants. We readily see that 

In the third quadrant the cosine is negative; 
cos 270° = 0. 

In the fourth quadrant the cosine is positive. 

Angles having Equal Cosines. If we suppose the angles 
XOm and XOM'^ (which have opposite algebraic signs) to 
become equal in absolute magnitude, the feet p and F" of 
the perpendiculars will coincide. Hence 

Tioo equal angles luith opposite signs have the same cosine. 

We may, by § 13, express this result in the symbolic forms 
cos y = cos (— y) — cos (360° — y), 
cos (180° + y) = cos (180° - y). 

We readily obtain this last formula by supposing the an- 
gles X'Oi¥and X'OM' to each become equal to y. 

We also have the theorem: 

To a given cosine may correspond two angles, whose sum 
is 0° or 360°. 



36 PLANE TRIOONOMETRT. 

41. Changes in the Tangent. The general definition of 
this function is: 

Hie tangent is represented hy the length which the terminal 
side of the angle cuts off from the tangent line N]Sf\ 

The tangent is positive or negative according as the ter- 
minal side intersects the tangent line on the positive or nega- 
tive side of X. We now see that the tangent will change as 
follows: 

As the line Om revolves round and m approaches Y, 
the point of intersection N will move 
upwards without limit. When m reaches 
Y, Om will become parallel to the tan- 
gent line, and the point N will recede to 
infinity. Hence 

The tangent of 90° is infinite. 

When m is in the second quadrant, 
the revolving side Om will not intersect 
the tangent line at all in the positive di- 
rection Om. We must therefore suppose 
the revolving line to be produced in the 
negative direction ON' so as to intersect the tangent line at 
N' below X. The distance XN' is then to be regarded as 
negative. Hence 

Li the second quadrant the tangent is negative. 

Following the motion, we see that when m reaches X', 
N' reaches X and the tangent becomes zero. Hence 

tan 180° = 0. 

When m is in the third quadrant, iV passes above X and 
the tangent is positive, so that 

In the third quadrant the tangent is positive. 

Also, noticing what takes place when m reaches P"', 

The tangent of 270° is infinite. 

In the fourth quadrant the tangent is negative. 

We see that whenever the terminal side moves through 
half a revolution it is turned round so as to cut the tangent 
line XA^in the same point as before. Hence 

To a given tangent correspond tiuo angles ivhose difference 
is 180°. 




FUNCTIONS OF UNLIMITED ANGLES. 



37 




42. Changes in the Cotangent The cotangent of any 
angle XOM is represented ly the Y 
length from Y to the point in 
which OM produced intersects 
YT. 

Following the point of inter- 
section as it moyes around the 
circle from X, we see 
cot 0° = 00 . 
The cotangent is positive in 
the first quadrant and diminishes from infinity to zero; 
cot 90° = 0. 
The cotangent is negative in the second quadrant and 
diminishes from zero to infinity; 

cot 180° = 00 . 
The cotangent is positive in the third quadrant; 

cot 270° = 0. 
The cotangent is negative in the fourth quadrant. 

43, Changes in the Secant. The secant is represented ly 
the distance OT from to the point N, in which the revolv- 
ing side intersects the tangent line XT. 

When m falls on X and the angle is zero, the secant is 
equal to OX, or unity. Hence 

sec 0° = 1. (3) 

As m moves from X to Y, the secant increases without 
limit and becomes infinite when m reaches Y. Hence 
sec 90° = 

As m moves from Y through 
X' to Y', the intersection of the 
revolving line with the tangent 
line falls in the negative part of 
OM, or in the direction of the 
dotted lines. Hence 

In the second and third quad- 
rants the secant is negative. 

At Y', when the angle is 
270°, the secant again becomes 
infinite. 

Between F' and X, or in the 
fourth quadrant, it is again positive. 



00 



(4) 




38 



PLANE TBIGONOMETBY, 



44. Changes in the Cosecant, The cosecant of XOM is 
represented hy the length of the side CM, intercepted letween 
and the line YT, 

Following the changes of the cosecant, we see that 
cosecO°=oo (because OX II YT). 

In the first quadrant the cose- 
cant diminishes from infinity to 
unity, remaining positive; 
cosec 90° = + 1. 

In the second quadrant the 
cosecant increases from unity to 
infinity, remaining positive; 
cosec 180° = 00 . 

In the third quadrant the cosecant increases from nega- 
tive infinity to negative unity, remaining negative; 
cosec 270° = - 1. 

In the fourth quadrant the cosecant diminishes from nega- 
tive unity to negative infinity, remaining negative. 

45. The algebraic signs of the several functions in the 
four quadrants are shown in the following diagram: 

Sine. Tangent. Secant. 





3 \^ ^4 

Cosecant. Cotangent. 



Cosine. 



46. Limiting Values of the Trigonometrical Functions, 

I. Sine and Cosine. The sine MP and cosine OP are 
necessarily not greater in absolute value than DM = 1. The 
limits of these functions are therefore -|- 1 and — 1. 

II. Secant and Cosecant. Since the tangent line lies 
without the circle, a secant can never be less than unity in 
absolute magnitude. But we have found that it may in- 
crease to infinity in either the positive or negative direction. 
Hence the limits of the secant and cosecant are 1 and infinity, 
and — 1 and — qo . 

III. Tangeiit and Cotangent. The limits of the tangent 



FUNCTIONS OF UNLIMITED ANGLES. 



39 



are easily seen to be — oo and + oo , or a tangent and cotan- 
gent may have any value wliateyer. 

47. When we know the numerical values of the sine, 
tangent and secant of all angles in the first quadrant, we may 
find the values of all six functions of any angle whatever, be- 
cause as we go around the circle the values of the functions 
are simply repetitions of the values between 0° and 90°. 

Let a be any angle less than 90°. Then any angle in the 
first quadrant may be represented by a. 

In the second quadrant it may be rejoresented by 90° -{- a, 
'' " '' '' 180° + «'. 

'' " " '' 270° -f a. 



'' third 
" fourth 



In the figure let XOM be any 



48, Second Quadrant. 
angle in the second quad- 
rant, and let us construct 
Angle XOM' 

= angle QOM 
= angle XOif- 90° 
= a. 
Then triangle OP'M' 
= triangle MP identically 
= triangle 0§Jf identically. 
Triangle OYN' 
= triangle OXN" identically. 
Triangle OYN'" 
= triangle OXiV^ identically. 

Comparing the equal lines which represent the six func- 
tions, we find : 

sin (90° + «) = 4- cos a; 
cos (90° + «') = — sin a; 
tan (90° -}- a) = - cot a; 
cot (90° -{- a) = — tan a; 
sec (90° -{- a) = — esc a; 
CSC (90° -\- a) z= -}- sec a. 

49. Third Quadrant. Comparing the equal lines in the 
diagram, and noting that the line OX, when considered as a 
side of the angle XOM = 180° + a^ is algebraically negative, 
we find 




From MP 



OP', 



OP = - M'P', 
XN = - YN"\ 
YN' = - XN'', 
OX = - OX'", 
OX' = OX", 



40 



PLANE TRIGONOMETRY. 




sin (180° + «) = - sin a; 
cos (180° -\- a) = — cos a-, 
tan (180° -\- a) =z -{- tan a; 
cot (180° -\- a) = -\- cot «'; 
sec (180° -\- a) = — sec «'; 
CSC (180° -\- a) = — CSC o'. 

50. Fourth Quadrant. Let 
F'Oi\^= XOif' = a. The fol- 
lowing triangles will then be iden- 
tically equal: 

OF'M' and MPO; OXiV^and OYN". 

Comparing the equal lines, we find 
PM = - OP' gives 

sin (270° + «f) = - cos a; 
OP = P'M' gives 

cos (270° + «) = + sin a; 
XN = - YN'' gives 

tan (270° -{- a) = - cot a; 
YN' = - XM'' gives 

cot (270° -\-a) = - tan a^ 
ON = OX'' gives 

sec (270° -\-a)=-^ esc a; 
OX' = - OM" gives 

CSC (270° -\-a) = — sec a. 

EXERCISES. 

1. From the values of the six functions of 30° in § 28, 
find the functions of 120°, 210° and 300°. 

2. Express the six functions of 105°, 195° and 285° in 
terms of functions of 15°. 

3. Prove the following equations by making the neces- 
sary constructions: 

sin (180° — x) = -\- sin x-, 

cos (180° — x) — —co^x-=— sin (90° — x)', 

tan (180° —x) = — tan x\ 

cot (180° — a;) = — cot a; = - tan (90° — x)-, 

sec (180° — x) — — sec x) 

CSC (180° - a:) = + CSC a; == + sec (90° - x). 




FUNCTIONS OF UNLIMITED ANGLES. 41 

sin (360° — x) = — sin x; 

cos (360° - ic) = + cos a; = + sin (90° — »); 

tan (360° - x) = — tan x; 

cot (360° - ic) =r _ cot a; = - tan (90° - x)] 

sec (360° - x) = -\- sec x-, 

CSC (360° — :?:) = — esc a; = — sec (90° — x). 

sin (270° - y) = - cos r; 
cos (270° -y) = - sin r; 
tan (270° - y) = -\- cot ;/; 
cot (270° -y) = -{- tan >/; 
sec (270° -y) — — CSC ;/; 
CSC (270° -y)=- sec )/. 

4. What relations subsist between the following pairs of 
functions? 

{a) sin (45° + a) and cos (45° — a)\ 
{h) sin (135° + a) and cos (135° - ay, 
{c) tan (225° + a) and cot (225° - a)-, 
(d) sec (315° + a) and cosec (315° - a), 

5. By means of the relations in § 25, deduce the following 
expressions for the value of each function in terms of each 
of the fiye others: 

tan a 1 



sin a = \/l — cos^ a 



Vl + tan^ a Vl + cof a 



y sec^ a — 1 1 



sec a 



cos a = yi -sm'a = 



cot a 



Vl + tan' a Vl + cot' a 



1 _ ycosec^ a — 1^ 
sec a ~ cosec a ' 



sin a Vl — cos^ <:jf 

tan a 



Vl — sin'« cos a cot a: 

= |/sec' a — 1 = 



|/cosec' a 



42 PLANE TRIGONOMETRY. 



pof; /v — 


Vl — sin' a 


cos a: 1 




sm a 

1 


VI — cos' a: tan a 





— Vcosec' a — 1 
1 




j/sec" (^ — 1 
1 


coo nr — 


— "■ — i/1 1 ti-n'' nr 




|/1 - sin^ a 


GOB a 




Vl + cot" a 
cot ^ 

1 _ 


cosec a 




Vcosec" a — 1 


noSPf^ CI — 


1 _ 1/1 + tan' « 




sin « |/i^ 


- cos' a tan a: 


__ 


-/l + cot' a 






|/sec' a — 1 



KoTE. In algebraic work of this sort the student will find it con- 
venient, instead of writing sin a, cos a, etc., in full, to use a single sym- 
bol for each function ; for example, he may put 



s = sin a; 


1 

— = cosec a 

s 


t = tana; 


J = cot a; 


c = cos a; 


1 

— = sec a. 

c 



ProYe the following relations 
6. 1 + sin X 



7. 



1 — sm X 
1 + cot _ tan + 1 
1 - cot ~ tan ^ - 1* 



8. sec 6 + tan & = - — ^ — - — ^. 

sec 6/ — tan 6 

9. {r cos xy + (r sin x sin w)' + (r sin x cos «^)'^ = r'. 

10. (a sin ;^)' + {a cos 7 sin d)' + (a cos 7 cos 6^)' — «'. 

11. (cos a cos Z* — sin ^ sin Z*)' 

+ (sin a cos b -{- cos o^ sin by = 1. 

12. (cos a -{- cos ;/)(cos a — cos ;/) 
= (sin / -{- sin «)(sin y — sin /ar). 



FUNCTIONS OF UNLIMITED ANGLES. 



43 




13. Using, if necessary, the figure in the margin, find four 
values of a: from the equation sin^ a=^, 

14. Find four yalues of x from 
the equation sin"" x = |. 

15. Of what four angles are the 
squares of the tangent each unity? 

16. From the conclusions of § 27, 
write the six trigonometric functions 
of 135°, 225° and 315°. 

17. From the conclusions of § 28, form the values of the 
following functions: 

Sines of 60°; 120°; 150°; 210°; 240°; 300°; 330°. 
Cosines and tangents of the same seven angles. 

18. For what two angles do we have tan a = 2 sin a? 

19. For what two angles do we have tan a = —2 sin a? 

20. For what four angles do we have tan^ a =z 4: cos^ a? 

21. Are there any angles for which the secant and cosine 
have opposite algebraic signs? 

22. Express the following trigonometric functions as func- 
tions of angles less than 45°: 



sin 50°; 


cos 50°; 


tan 50°; 


cot 50°. 


sin 130°; 


cos 130°; 


tan 130°; 


cot 130°. 


sin 140°; 


cos 140°; 


tan 140°; 


cot 140°. 


sin 245°; 


cos 245°; 


sec 245°; 


CSC 245°. 


sin 335°; 


cos 335°; 


sec 335°; 


CSC 335°. 



23. From the conclusion of § 15, show that if we increase 
an angle by any entire number of circumferences the values 
of the trigonometric functions will all remain unchanged. 

24. As an example of the preceding theorem prove the 
equations 

sin 370° = sin 10°; 
sin 350° = sin (-10°). 

25. Express as functions of angles less than 45°: 

sin 862°; cos 593°; tan 790°; cot 1020°. 

26. In what quadrants do we have 

sine and cosine both negative? 
sine and cosine both positive? 
sine negative and cosine positive? 
sine positive and cosine negative? 



44 PLANE TRIGONOMETRY. 

27. If the sine of an angle be given, how many values may 
the cosine have? 

It is desirable to find both a geometric and an algebraic proof: the 
latter from § 25. 

28. If the cosine of an angle be given, how many values 
may the secant have? 

29. Find two angles of which the sines are double the 
cosines. 

30. Find two angles of which the cosines are double the 
sines. 

31. If sin 2/ = 3 cos y, and cos a; = 3 sin x, show that 
a; + «/ = 90° or 270°. 

32. Show between what limits the sum of the sine and co- 
sine of an angle is positive, and between what limits negative. 

33. Find two values of the angle x from each of the equa- 
tions 

sin X -\- Qo^x = ^', 
sin X + cos cc = — I". 

34. Find, two values of x from 

sec X + tan x = ^. 
Eeduce the following expressions to their simplest form: 

35. a cos (90° + .) + — g^. 

36. m sin (180° - x) -\- n sin (180° + x). 

37. cos X (cos X — sec x) + ^^^"^ (180° — x). 

38. tan (180° + x) tan (180° - x) + sed" x. 

39. sin' c^ {1 + tan' (90° + a)\. 

40 . 

1 + tan' « ^ 1 + cot' a 

41. (m + n) sin (180° -j- a) - {m - n) sin (180° - a). 

42. (m + n) cos (180° + a) - {m - n) cos (180° - «). 

43. (m — n) sin x -\- (n — p) sin (180° — x) 

+ {p - m) sin (180° + x). 

., a (1 — cos oi) (1 4- cos O') , , , , „ 

44. -7^ ^ — , '^ , . ~ -\- a — I — I cot' a. 

(1 — sm a) (1 + sm a) 

45. {p + q) cos (270° - «) + (^ + r) cos (270° + a) 

+ (r +i?) cos (90° -\-a), 

46. (r + 8) sec « + (r — s) esc (90° + a) 

- (r + s) sec (180° ■\- OL) -{j- - s) CSC (270° + a). 



CHAPTER V. 

RELATIONS BETWEEN FUNCTIONS OF SEVERAL 
ANGLES. 



The Addition and Subtraction Theorems. 

51. Peoblem I. To express the sine and cosine of the 
sum of two angles in terms of the 
sines and cosines of the angles them- 




•1 



Solution. Let 

XOM=a ^ndi MON= ^ 
be the two angles. XON = <af + y5 is 
then their sum. 

Let ON be the unit radius. From 
i\^drop 

NMiOM', NQlOX. 
From M drop 

MP 1 OX; MR 1 NQ. 
Then sin {a -\- ft) = NQ = NR -^^ MP', 

cos {a-^ 13)= 0Q= OP - MR. 
OQS and XMS being right angles, we have 

Angle RNM= comp. RSM. = comp. OSQ = 80 Q = a, 
By §31, 

NM= sin /?; 

NR = iYif cos « = sin yS cos a; 
OM = cos /3; 

MP = OM sin a = cos /? sin a; 
whence, from (1), 

sin {a-\- j3) = sin a cos j3-\- cos a sin jS, 
We find in the same way 

OP = OM cos a = cos J3 cos a; 

RM== iOfsin a = sin /3 sin a; 

whence, from (1), 

cos (a -\- ft) = cos a cos ft — sin a sin /?. 



(1) 



(3) 



(3) 



46 



PLANE TBIQONOMETBT. 




The formulae (2) and (3) constitute the addition theorem 
of trigonometry. 

Pkoblem II. To express the sine and cosine of the dif- 
ference of two angles in terms of 
the sines and cosines of the angles 
themselves. 

Let POM be the angle a, and 
i\^Oif the angle/?. Then 
PO]Sf=^ a -13. 
Take ON for the unit radius 
and drop 

NP 1 OP; NMl OM', 
MQiOP; JSTPiMQ; 
Then sin (o(,~ J3) = PN = QP = MQ - ME; (a) 

cos {a - J3) = OP = OQ + RN. (b) 

ecause OMN is a right angle, 

NME = comp. OMQ = MOQ = a; 
MN-=z sin ^; 
OM = cos /?; 

MP = MN cos NMR = sin ^ cos a; 

MQ.= OM sin MOQ = cos /5 sin «; 

OQ = OM cos JfOC = cos ^ cos or; 

PJV = JO^sin JVMP = sin /3 sin «f. 

Making these substitutions in (a) and (b), we have 

sin («r — y^) = sin a cos /? — cos a sin /?; (4) 

cos {a — J3) = cos flf cos ^ + sin a sin /?. (5) 

Remaek. In the constructions we have drawn the figures as if 
a and /? were each less than a right angle. But the same reasoning 
will apply to constructions in which these angles have any value what- 
ever, provided that we assign the proper algebraic signs to the lines 
which enter into the demonstration. The work necessary to this ex- 
tension will be a valuable exercise for the student. In constructing 
each function he must have recourse to the general definitions in 
§§ 38-40. 

52. Sine and Cosine of Twice an Angle. If we suppose 
j3 = a, we have from (2) and (3) expressions for the sine and 
cosine of the double of an angle, namely: 

sin 2a = sin a cos a -{- sin a cos a, (6) 



RELATIONS BETWEEN FUNCTIONS OF ANGLES. 47 

or sin 2a = 2 sin a cos a\ (6') 

cos 2a = cos'^a — sin^^ = (cos a -f- sin a) (cos a — sin a). (7) 
Also, by putting for cos' a its yalue, 1 — sin' a, and vice versa, 
we have 

cos 2a = 1-' 2 sin' «a: = 2 cos' <a? — 1. (7') 

EXERCISES. 

1. Because 90° = 60° + 30°, we liaye, by putting a = 60° 
and /3 = 30° in the equation (2), 

sin 90° = sin 60° cos 30° + cos 60° sin 30° = 1. 
It is required to test this equation by substituting the nu- 
merical yalues of the sines and cosines of 30°, 60° and 90° 
(§ 28), and to test in the same way the equations obtained by 
putting a = 60° and j3 = 30° in (3), (4) and (5). 

2. Because a = a — x -\- x, we have, from (2), 

sin a = sin {a — x) cos x -\- cos {a — x) sin x. 
It is required to write the corresponding equations ob- 
tained by making the same substitution in (3), and the equa- 
tions obtained in the same way from (4) and (5) by the iden- 
tical equation 

a = a -{- X — X. 

3. Derive the addition theorem for the cosine from that 
for the sine by substituting y — 90° or 90° — y for /? in the 
equation (2), and applying the equations of § 22. 

4. By means of the addition theorem prove the equations 
sin (a -\- /3 -}- y) = sina cos /3 cos y -{- cos a sin /3 cos y 

+ cos a cos /3 sin y — sin a sin ^ sin y; 

cos ((X -\- J3 -{- y) = cosa cos j3 cos y — sin a sin /? cos y 

— sin a cos J3 sin y — cos a sin J3 sin y. 

Note. This is readily done by putting a -{- /3 tor a, and y for j3, 

in the equations (2) and (3), thus giving 

sin {a-{- ^ -{- ^) = sin (a -\- /3) cos y + cos {a -\- /3) sin y, 
cos {a -{- /3 -{- y) = cos {a -\- /3) cos y — sin {a -}- /3) sin y, 

and then developing by the addition theorem. 

5. Prove the following values of sin da and cos 3a: 

sin da = d sin a cos' a — sin^ a; 
cos 3 a = cos' a — 3 sin' a cos a. 



48 PLANE TRIGONOMETRY. 

6. Transform the expression 

a cos {a-\-x) -\-h cos (/? + a;) + c cos {y + ^) 
into 

cos X {a cos a -^-h cos /? + c cos y) 
— sin aj (a sin or + 5 sin /? + ^ sin y). 

7. If we have 

a cos or + 5 cos /? + c cos / = 
and a mn a -{- d ^m 13 -\- c Bin y — 0, 

prove that we must also have 

a sin {a-{-x) -{-h sin (^ + a?) + c sin {y -\-x) = 0, 
whatever be the value of a:. 

53, The Addition Theorem for Tangents, Dividing 
equation (2) by (3), we have 

mi {a -\- 13) _ . j^ a\ _ sill ^ cos /? + cos a sin p 
cos {a -{- ft)~ ^ T H) — ^Qg ^QQgj^_ g^-Q^ ^ gjjj y;5' 

Dividing both numerator and denominator of the last mem- 
ber by cos a cos /3, the equation becomes 

, , , „. tanor + tan/J 
^'>'^('^ + ^>= l-tan«taix/? - ^^^ 

We obtain in the same way, from (4) and (5), 

, , ^. tan « — tan /3 ,^. 

Putting fi = am (8), we have 

. ^ 2 tan Of .... 

tan 2ar = — —, (10) 

1 — tan a ^ ' 

EXERCISES. 

1. Assuming p = 180°, prove by (8) of this section that 

tan {a + 180°) = tan a. 

2. Assuming a = 30°, substitute in (10) the value of tan 
30° given in § 28, and thus obtain the value of tan 60°. 

54. Products of Sines and Cosines. Taking the sum 
and difference of equations (2) and (4) of § 51, and reversing 
the members of the equation, we find 

2 sin a cos p = sin {a -{- j3) -{- sin (a — /3);) 
2 cos a sm /3 = sin (a -\- P) — sin (a — /3), ) ^ ^ 
In the same way we find, from (3) and (5), 

2 cos a: cos ^ = cos {a -\- P) -\- cos (a — /3); ) 

2 sin a sin ^ = - cos (a -{- p) -{- cos (a-p). \ ^^^^ 



(13) 



BELATIONS BETWEEN FUNCTIONS OF ANGLES. 49 

EXERCISES. 

1. Prove that ii a - /3 = 90°, 

then cos (^ -]-/?)= 2 cos acos ^ = sin 2a = — sin 2/?. 

2. Proye that if a -{- J3 = 180°, 

then sin (a — fi) = 2 sin acos /3 = sin 2/3 = — sin 2af. 

55. /S^-wm 0/ /Sme^ and Cosines. If in the four equations 
(11) and (12) we put 

a -\- /3 = X 
and a — ^ = y, 

we shall have 

and p = i{x — ?/). 

By substituting these values in (11) and (12), and inter- 
changing the members of the equation, we find 

sin X -}- siny = 2 sin -l{x + y) cos ^(x — y) ; 

sin X — sin y = 2 cos ^(x -\- y) sin ^{x — y); 

cos ic + cos y = 2 cos ^{x -\- y) cos J-(:z; — y) ; 

— cos a; + cos ^ = 2 sin ^{x -\- y) sin ^(rc — y), , 

Dividing the first of this group of equations by the third, 

we get 

sin a; + sin ?/ , w , x ,., .x 
^ = tan Ux + y), (14) 

cos X + COS 2/ rf \ ' C7 / \ / 

Dividing the second by the fourth, we get 

sin a; — sin ?/ . w . a /-. k\ 
^ = cot Ux 4- v). (15) 

cos ^ — cos a; ^\ • ;?/ \ / 

If in the last two equations of (13) we suppose y = 0, 
which makes cos y = 1, they become 

1 + cos a; = 2 cos^ ^x, ) . . 

1 - cosa; = 2sin'ia;; j ^16) 

a pair of equations which frequently come into use. 
If we put a = ^x, these equations become 

1 + cos 2<a: == 2 cos'^ a, ) naf\ 

1 - cos 2a: = 2 sin'' a-, \ ^^^ ' 

which may also be derived directly from (7'). 

Dividing the second of (16) and (16') by the first, we find 

1 — cos iC , 2 1 /-, ^x 

rT"c^ = *"''*''' ^"^ 

l^°i|f = tan^,. (170 

1 + cos 2<a: ^ ^ 



50 PLANE TRIGONOMETRY. 



EXERCISES. 



i. Prove that the functions of half an angle are given by 
the equations 

. . a/1 — cos a 

.a = y - 



sm 



2 



./I + cos a 



cos.„_. ^ 

, ^ sin a 

tan far 



cot ^a = 



1 4" cos a' 
sin Of 



1 — cos or 
Deduce the following equations: 

o t I 4- /? sin (« + /?) 

2. cot a -{- Got fi = -^ — ^-h^' 

sm a sm p 

ox I X /? sin (a + /5) 

3. tan or + tan /? = ^^ — —^* 

cos a cos p 

^ -I 4. 4. /? cos (or + /5) 

4. 1 — tanatany5= ^^ !~^-. 

cos a: cos p 

e -I . X i- /? sin (« + /5) 
5.1 + cot or tan /? = -r— ^^ — —hf' 
' sm a: cos p 

n J. 1 a COS (a: + /?) 

6. cot « — tan y? = -^— '^ '— ^-. 

sm «: cos p 

^ sin (:» + ?/) _ tan x -\- tan ^ _ cot y -\- cot a; 
sin {x — y) ~ tan a; — tan y ~ cot y — cot a;* 
cos [x -\- y') _ cot ^ — tan x __ cot a; — taii y 
cos (x — y) ~ cot ^ + tan x ~~ cot cc + tan y 

9. sin'^ « cos^ /3 — cos* « sin'* ^ = sin (<af + P) sin (« — fi)- 

10. cos'' ^cos^/? — sin'* a sin'* /? = cos {a -\- /3) cos (a — /3). 

11. sin (d + q)) cos — cos (6^ + (p) sm ^ = sin <p, 

12. 2 sin a sin (90° — «) = sin 2a. 

13. 2 sin (45° + a) sin (45° - a) = cos 2«. 

14. 2 cos (45° - a) sin (45° + a) - 1 = sin 2a. 

^ ^ . . 2 tan ^A 

15. sm ^ = ., , . — 77-:-. 

l+tan'*^^ 

2 

16. sin A = -7 — 7-7-7 r-TT' 

tan -|-^ + cot -J-J. 



8. 



RELATIONS BETWEEN FUNCTIONS OF ANGLES, 51 

. , sin (30° -\-A)- sin (3 0° - A) 

17, sm A = ^ . 

18. sin ^ = 1 - 2 sin^ (45° - i^). ^ 

1 — tan^ Ja 
19- ''"^ " = l + tau'ia - 

coHa — tan-|« 

21. cos a =2 cos (45° — Ja) cos (45° + ^a). 

22. sec a =i tan (45° + i^) + i tan (45° - ia). 

23. sec flj = 1 + tan a tan -1^?. ..—- 

. r. 2 cot id 

^^' *^^ ^ = cotH^^' 
, ^ sin 2^ 

25. tan = :^ — [-——Kn' 

1 4- cos Zu 

26. 2 tan 6 = tan (45° + id) - tan (45° - id), 

sin a; , sin x 

27. Show that the expressions ^^^- and ^ _ ^^^ ^ 

are reciprocals of each other. 

28. From the equations (13) express tan i{x — y) and 
cot \{x — y) in terms of the sines and cosines of x and y. 

ProYC the following equations: 

39. \±^ = i(l + tan «)^ 

1 + COS 2« 

tan 6 
30- t^'i i^ = 1 + sec e - 
31. cos (30° - a) - cos (30° + «) = sin or. 
33. cot (30° + e) - tan (30° + 6") = 2 tan (30° - 20). 

3 cos 20 + 1 

33. cot (30° + d) cot (30° -6) = ^^^^Jq^TT 

34. tan 40° + cot 40° =: 2 sec 10°. 

35. ?»^^"-tan}«. 
COS a -\- COS 2a 

36. cos « - cos 2^ ^ t^^ ^^ tan |^. 
COS a + COS 2a 

37. COS acos /3 = cos' i(« + Z^) - sin' i{a - /?). 

38. sin a sin /? = cos' i(« - ^) - cos'^ i{a + /?). 

1 1 

39. cot (« + r) = tan a + tan ^ cot a + cot y 



52 PLANE TRIGONOMETRY. 

40. tan^ a — tan^ fi = sin {a + /?) sin {a — /?) sec' a sec' fi. 
If or, /? and y are in arithmetical progression, prove the 

following four equations: 

41. sin a: — sin ;/ = 2 sin {a — J3) cos /?. 

42. cos Of — cos ;/ = 2 sin (y5 — a) sin /?. 

43. i(cot /3 - tan >5) = \-- — ] — -. 

^ ' ' ' tan a + tan y cot or -|- cot y 

, , , , ^. cos a — cos y sin y — sin a 

4:4:. tan (a — fi) = -r- ■ — r—^ = ^— . 

"^ ' '' sin <af -f sm ;^ cos y -j- cos a 

Prove for all values of the symbols: 

45. (cos a — cos /Sy -j- (sin a — sin jSy = 4 sin" i(a — /3). 

46. cos(^ + ^+ r) ^ ^ _ ^^^^ ^^^^ _ ^^^p ^^^ 
cos a cos p cos y / / / 

— tan ;/ tan or. 

— cot a — cot y^ — cot y. 



.„ cos(ar + ^-|-v) , , . . 

47. -T — ^ .' ^ ! ^^ = cot or cot 6 cot v 
sm a sm p smy ' ' 



48. cos 2a = ^ — r—z — ^-• 

1 + tan a 

49. cos' « — cos' Sa = sin 2a sin 4«. 

50. (1 + cot a -\- CSC a) (1+cot a — esc a) =. cot ^a —tan ^a, 

51. By combining the equations (8) and (10) of § 53 find 
tan 3 Of in terms of tan or. 

52. From (10) find tan ia in terms of tan a, 

53. From § 27 find the six functions of 22^° and of 67i°. 

54. If tan' 6 = tan (a - 6) tan {a + d), 

then sin 2^ = V2 sin a. 

55. If sec {(p-\- a) -{• sec {cp — a) =:2 sec cp, 

then cos cp = l^cos ia. 

56. If tan i(a + /3) tan |(«r - fi) = tan' ^y, 

then cos of = cos /3 cos ;^, 

57. If cot 2d — — tan q>, then tan (6 — <p) z= cot 0. 

58. If tan a; = J, and tan y — ^, then sin {2x + t/) = —=. 

59. Prove: 

(p + ^) tan (315° -[- oc) - {p - q) tan (315° - «^) 
-{p -{- q) tan (225° + «) + (V - ^) tan (225° - a) 
= — Aq sec 2a, 



CHAPTER VI. 

SOLUTION OF TRIANGLES IN GENERAL. 



56. A plane triangle lias six parts, three sides and three 
angles. Of these parts the three angles are not independent 
of each other, because when two angles are giyen the third 
may be found from the condition that the sum of the three 
angles is 180". Hence, if two angles are given, the case is 
the same as if all three were given. 

When any three independent parts are given, the remain- 
ing three may be found, but in order to be independent one 
at least of the three given parts must be a side. 

The given parts may therefore be 

One side and the angles; 
Two sides and one angle; 
The three sides. 

Also, when the sides and one angle are given, this angle 
may be either that between the given sides or opposite one of 
them. Hence there are four cases in all to be considered. 

Remark. 

§74. 



When two angles are given, the third is found by Geom., 



5 T • The fact that the sum of the three angles of a plane 
triangle is 180° enables us to express a trigonometric function 
of any one angle as a similar function of the sum of the other 
two angles. It has been shown that 

sin X = sin (180° — x 

cos a; = — cos (180° — X 

tan a; = — tan (180° — x 

cot X — — cot (180° — X 

sec :?; = — sec (180° — x 

Q^QX — — GSC (180° — X 



54 



PLANE TBIOONOMETRY. 



If a, /? and y are the three angles of a triangle, we haye 



a = 180^ 
/? = 180^ 

r = 180^ 



(« + /?). J 



Therefore (§§ 38-40) 

sin «r = sin (^ -\- y); 
sin /? = sin (y -J- a:); 
sin y = sin (« -j- /5); 
cos « = — cos (y^ -|- y);^ 
etc. etc. 

By diyiding the equations (a) by 2 we find 

i« = 90° - i(^ + r)n 
i^ - 90° - Ur + ^); [ 
ir = 90° - i(« + /5). J 



Therefore (§ 21) 



sin-|«: = cosi(/? + v)'," 
cos -Jar = sin |(^ -f ;/); 
tan !«: = cot -|(/? + ;/); 
cot ^a = tan |^(y5 -\- y)'y. 
etc. etc. 



(«) 



(1) 



{^) 



(2) 



58. Case I. Criven the angles and one side. 

Theoeem I. The sides of a plane triangle are propor- 
tional to the sines of their opposite angles. 
Proof Put 
a, l, c, the three sides; 
a, p, y, their o|)posite angles. 
From one angle, as y, drop a per- 
pendicular yD upon the opposite 
side, c. Then 

yD = h sin «f ; yD = a sin /?. (§35) 
Therefore 

a^m f3 — 1) sin a, 
Diyiding by sin a sin fi, 

a _ t 
sin a sin /J' 




SOLUTION OF TRIANGLES IN GENERAL. 55 

By dropping a perpendicular from a upon a we should find, 
in the same way, 

_b____o_ 
sin /? " sin y 

r„, « ah c ,^. 

Therefore ~. = -^— ^ = -. , (3) 

sm a sm p sm y ^ ^ 

or a : 5 : c = sin a: : sin /? : sin y. Q.E.D. (4) 

Def. The common yalue of the three quotients -^ , 

•^ ^ sm a' 

t c 

-. — ^ and is called the modulus of the triangle. 

sm p sm y ^ 

Theorem I. enables us, when the angles and one side of 
any plane triangle are giyen, to find the remaining two sides. 
If the side given is c, we haye 



csm a 

a= ~. ; 

sm y 

, _ c sin y5 



(5) 



sm y 

Or, putting ilffor the modulus, we haye 

a= if sin «; "] 

h--= Msm fi',V (6) 

c = M ^m y. \ 

If we put^, p' and j^j" for the lengths of the perpendicu- 
lars from a, f3 and y respectiyely upon the opposite sides, 
we find, from the preceding figure, 

p = 5 sin ;/ = c sin /?; "I 

p' = c mi a = a sin y; \- (7) 

p''= h sm a — a sin /?. J 

By these equations we may find the lengths of the perpen- 
diculars. 

EXERCISES. 

1. Giyen a = 78° 23', /3 = 52" 16', a = 796.2; find d and c. 

2. '' a= 5° 26', /? = 72° 36', 5 = 19.26;** a and c. 

3. '' a = 50° 58', p = 32° 50', c = 169.3; '' a and d. 



PLANE TRIGONOMETRY. 



B 



4. In order to find the distance of a point G across a river 
from the points A and B, a 
surveyor measured a base line -^r--- _ 
AB and found it to be 829. 72 """ ||| r ----. 

metres. Placing a theodolite 1 |||||j •HHIIl-^---V-'C 

at ^, he found 

Angle ^^(7= 82° 37'. 6. 
Carrying his theodolite to B, 
he found 

Angle ^^C= 70°3'.3. 
Required the distances ^4Cand BC across the river, and the 
length of the perpendicular from (7 on AB. 

5. In a triangle ^^Cthe angle B exceeds the angle A by 
10°, the angle C exceeds the angle B by 20°, and the side A G 
is 2. 7294 metres. Find the angles and sides of the triangle 
and the length of the perpendiculars from the angles upon 
the opposite sides. 

6. The base of an isosceles triangle is 132. 64 metres, and 
the angle at the vertex is 32° 53'. 7. Find the sides and the 
altitude. 

7. One diagonal of a parallelogram measures 23 metres, 
and it makes angles of 32° 17' and 63° 24' with the sides. 
Find the sides and angles of the parallelogram. 

8. The angles at the base of a triangle are 42° 25' and 58° 
32', and the sum of the sides opposite these angles is 52. 78 
metres. Find the three sides. 



KoTE. If a and b are the sides, we have 
« + 5 = 52.78, 



b sin y3' 
from whicli a and b may be found. 

9. The perpendicular upon the base of a triangle is 32.52 
metres, and the angles at the base are 62° 12' and 75° 28'. 
Find the three sides. 

10. The two parallel bases of a trapezoid are 72 feet and 18 
feet, and the angles at the ends of the longer base 80° 5' and 
52° 47'. Find the non-parallel sides and the area. 



SOLUTION OF TRIANGLES IN GENERAL. 



57 



11. Show that lip be the perimeter of a triangle, we shall 
have 

P 



Modulus 



sin a + sin /5 + sin ;^ 




12. Two of the three angles of a triangle are 22° 50' and 
78° 20' respectively, and the perimeter is 226 metres. Find 
the three sides. 

13. The perimeter of a triangle is 100, and the angle A 
exceeds the angle B by 15° and the angle G by 25°. Find 
the three side^. 

14. The base of a trapezoid is 24.12, the altitude is 5.76, 
and the angles at the base 
are 65° 22' and 59° 36' re- 
spectively. Find the three 
remaining sides. 

15. In a trapezoid 
ABGD are given: angle 
A = 53° 55', angle 
B = 82° 12', angle BAG 
made by one diagonal and 
the base = 24° 27', and the 
length ^C=: 59.22. Find 
the remaining angles and 
the sides. 

16. One diagonal of a parallelogram makes angles of 
101° 20' and 33° 35' with the sides, and the sum of this diago- 
nal and one lesser side is 32.6. Find the sides of the paral- 
lelogram. 

17. The base of a parallelogram is 37.7, and the diagonals 
form with it angles of 30° 25' and 42° 20' respectively. Find 
the lengths of the diagonals. 

18. The perimeter of a parallelogram is 2434, and one 
diagonal makes angles of 35° and 45° respectively with the 
sides. Find the sides. 

19. The angle at the vertex of a triangle is 52° 12', the 
greater angle at the base exceeds the lesser by 12°, and the 
one side exceeds the other by 5.23. Find the sides and the 




B 



58 PLANE TBIGONOMETBT. 

59. Case II. Given two sides and the angle 
opposite one of them. 

Let the given parts be a, h and a. We then compute the 
parts fi, y and c by the formulae {a) and (3) already found. 

• /? ^ • 1 

sm /) = — sm a-, 

K = 180° - (« + /J); !- (8) 

_h^my _amiy 
~ sm/3 ~ sin a: * J 
This case may have two solutions (Geom., §§ 111, 113). 
The two solutions are found in the above equations, because 
to a given value of sin J3 corresponds either of two angles /3 
(§39), which will be supplements of each other. 

But if one solution gives a-\- /3 > 180° it is not admissible, 
and only the lesser value is used to give the triangle. This 
will be the case when a > l. 

It may also happen that sin ^ = — sin a comes out greater 

it 

than unity. There is then no possible triangle which fulfils 

the conditions. 

Example. Given a = 152.08, b = 236.74, a = 32° 30'; 

find the remaining parts. 

log &, 2.3742 

log sin «', 9.7302 

co-log a, 7.8180 



log sin/?, 9.9224 
/3, 56° 46 
a, 32° 30 


or 

or 
or 

or 
or 


123° 14' 
32° 30' 


« + A 89° 16 
180° 00' 


155° 44' 

180° 00' 


y, 90° 44 
log sin y, 0.0000 

log -X-,, 2.4518 
^ sm /5' 


24° 16' 
9.6138 

2.4518 


log c, 2.4518 
c, 283.00 


2.0656 
116.30 



SOLUTION OF TRIANGLE8 IN GENERAL. 59 







EXERCISES. 






1. 


Giyen 


« = 24, J = 33, a= 31° 28'; 


find rem. 


parts. 


2. 


iC 


a^U,!) = 35.79,y5 = 17° 59'; 


<( 


a 


3. 


a 


^ = 29, Z* = 34, « = 30° 20'; 


a 


<e 


4. 


i( 


5 =: 19, c =3 18, y = 15° 49'; 


(C 


t{ 


5. 


(( 


a = 24, c = 13, a = 115° 0'; 


iC 


(( 



6. From a point P at a distance « from the centre of a 
circle of radius r a line is drawn, making the angle /3 with 
the line from P to the centre of the circle. At what dis- 
tances from P will the line intersect the circle? Compute 
the distances when r = 72, « = 98 and j3 = 28° 56'. 

Case III. Given the three sides. 

60. Theoeem III. In a triangle the square of any side 
is equal to the sum of the squares of the other two sides minus 
tioice the product of these two sides into the cosine of the angle 
included ly them. 

In symbolic language this theorem is expressed in any of 
the forms 

c^zzil^ -\.(? — %bc cos a, \ 
or y = 0" + c'^ - ^ac cos /3, V (9) 

or c^ =za^-\-y^ — %al) cos y. ] 

Proof In any triangle the square of any side is greater 
or less than the sum of the squares of the other two sides by 
twice the product of one of these sides into the projection of 
the other side upon it (Geom., § 312-314). 

If a be any side, we have, by this theorem, 

a^ = If ^ d^ ^ 21) X (projection of c on I). 

By §36, 

Projection of c on Z* = c cos a. 

Now cos a is positive or negative according as a is acute 
or obtuse (§ 40). When the angle a is obtuse, the expression 
25c cos a being negative, — 2Jc cos a will indicate the numer- 
ical addition of the product 25c cos a, as required by § 313 
of Geometry. 

When angle a is acute, 25^ cos a being positive, ^Ic cos a 
will indicate the subtractior of the product %ic cos a, as re- 
quired by § 312. 




60 PLANE TBIQONOMETMT. 

Hence, if we have regard to the algebraic sign of c cos a, 
we may always write 

a' = Z»' -f c' — ^hc cos a, Q.E.D. 

The other equations may be proved in the same way. 

We may also, from any one general 
equation between the sides and angles of 
a triangle, obtain two other equations of 
the same kind by supposing the sides and 
angles arranged in regular order around a 
circle, and substituting for each side the 
side next following, and for each angle the 
angle next following, as shown in the fig- 
ure. The reason is that any proof which is valid for any sides 
and angles will be equally valid when applied to the sides and 
angles which follow them in circular order. 

Thus we may derive the second and third of equations (9) 
from the first by a simple change of letters. 

61. From the first of equations (9) we obtain 

cos a = — ^ — nrw 

which, with the two companion formulae, enable us to find 
the angles when the three sides are given. 

If the angle is small, it cannot be accurately determined 
by means of its cosine; we therefore transform the expression 
as follows : 

Subtracting each member of the equation from unitv we 
have • 

1 - cos «f = 1 - ^'+o'-ct' ^ 2bc-h^-c^-^ a^ 

But 

1 — cos or = 2 sin' \a (8 55), 
and ^^ ^ 

%bc - 5' - c' = - (5 - c? 
Therefore ^ ^ 

2 sin«i« = a'~{l-cr ^ {a^ l~c){a^l-^c) 

Let us now put s for half the sum of the three sides, so 
that 



SOLUTION OF TBIANGLE8 IN GENERAL ,61 

Then a -]- i — c = 2s — 2c] 

a — l -\- c = 2s — 21', 
and the preceding equation reduces to 

sinH«^ = ^^ Y^ -' (11) 

The expressions for the other two angles are obtained by 
the same process, the letters a, l, c and a, p, y being per- 
muted in the orders J, c, a\ fi, y, a and c, a, h ; y, a, /3. 
We thus find 

sin-i/^^(---; ^---) ;| 

'''' ^^ - ah -J 

62. The equations (11) and (12) answer our purpose, but 
in determining an angle the tangent is the function to be pre- 
ferred, because an angle can be determined more accurately 
from its tangent than from its sine or cosine. To obtain 
expressions for the tangent add unity to both sides of the 
equation (10). We then have 

, , &' + c' - «' ^' + ^^c -\- c'-a' 
1 + cos « = 1 + ^-^^— = -^^^ . 

Since 1 + cos a — 2 cos' ^a (§ 55), this equation reduces to 
^ cos ^a - ^- - —^^ . 

Whence cos^Jo: = ^^^-^^. (13) 

Dividing (11) by this equation, and writing the corre- 
sponding equations for the other angles, we find 

(s - S)(« - c) 



'^ s(s — 0) 

tan 2r - ^(^ _ ^) • 



(14) 



The computation will be simplified a little by computing 

E=: |/ ('^~^-)(^~^)(^~^) . 



62 



PLANE TBIOONOMETRT. 



We shall then have 



tan ^a = 
tan iy5 = 
tan ^y = 



H 



{s-aY 
H 

{s-hy 

H 



(15) 



{s - c)' 

By means of these equations we may compute two of the 
angles, and find the third by subtracting their sum from 180°. 
But in practice it is better to compute the three angles inde- 
pendently^ and check the accuracy of the work by taking 
their sum. 

If this sum comes out materially different from 180°, there 
is some mistake in the work; if not, it may be presumed cor- 
rect. 

Example. Given a = 273.96, b — 198.63, c = 236.91; 
find the angles. 



a = 273.96 
b = 198.63 
c = 236.91 
2s = 709.50 
s = 354.75 
80.79 
156.12 
117.84 



a = 






log (5 - a), 1.9073 

log {s - h), 2.1934 

log (s - c), 2.0713 

sum of logs, 6.1720 

log s, 2.5499 

log H\ 3.6221 

log IT, 1.8111 

-la, 38° 42'; a = 77^ 
■l-A 22° 31'; /3 = 45^ 

ir, 28°47'; r= 6r 



log tan ^a, 9.9038 
log tan I/?, 9.6177 
logtsLn^r, 9.7398 



24' 

2' 

34' 



Sum = 180° 0' (Check.) 

Note. The check may come out one or two minutes in error from 
the imperfections of the four-place logarithms. 

Another check on the accuracy of the work is obtained by 
computing the modulus of the triangle from its three separate 
expressions (§ 58, 3), and noting whether they agree, thus: 
log a, 2.4377 log d, 2.2981 log c, 2.3745 

sin or, 9.9894 sin/?, 9.8497 sin >/, 9.9263 



log modulus, 2.4483 



2.4484 



2.4482 



SOLUTION OF TRIANGLES IN GENERAL. 63 

The three results agree within the unavoidable limits of 
error. 

EXERCISES. 

1. Given a = ^, '6=4:, c = 5; find the angles. 

2. '' a = 37593, 5 = 29867, c = 40005; '' 

3. '' a=: 2.7961,5 = 23.928,c = 25.046; '' 

4. The base of a parallelogram is 13, each side is 6, and its 
lesser diagonal is 12. Find its angles and its greater diagonal. 

5. If the sides of a parallelogram are a and I?, and the 
diagonals are p and q, express the cosines of its angles. Then, 
by the relation which must exist between the cosines, prove 
the theorem that the sum of the squares of the diagonals is 
equal to the sum of the squares of the four sides. 

6. The parallel sides of a trapezoid are 12 and 17, and the 
non-parallel sides 6 and 7. Find its four angles. 

Suggestion. Divide the trapezoid into a triangle and a parallelogram. 

7. In a triangle are given the two 
sides p and q and the medial line r vy^ I \? 
from the vertex to the middle point 
of the base. Prove 




= 1/2^4- 2^=^- 4rl 

8. G-iven the three medial lines r, r', r" of a triangle; 
find the sides of the triangle from the preceding result. 

Ans. I |/2r^+2/^-r"=^; f 4/2r^+2r"^-r'^; f |/2r'^+2r"^-r^ 

9. Of the bisectors of the angles at the base of a triangle, 
the one cuts the opposite side in the ratio m : n, and the other 
in the ratio p '. q. By means of the 
equation (10) express the cosine of ty^\q 
the angle at the vertex of the tri- 
angle. 

Ans. cos a = !— ^ ~ . 

Zpqmn 

Note, In the solution of this question apply § 405 of Geometry. 

10. The perimeter of a triangle is 37 metres, and its three 
angles are 82°, 54° and 44° respectively. Find its three sides. 

Note. Begin by applying the theorem of Geom., § 371, to the pro- 
portion (4) of § 58. 




64 PLANE TBIGONOMETRT, 

11. Two men start from the same point. A goes west 32 
kilometres, and B goes west 40° south until he is 20 kilometres 
from A's stopping-point. What is the distance he has tra- 
yelled? Find two answers. 

12. From a centre distant 22 metres from the vertex of an 
angle of 25°, on one of its sides, and with a radius of 18 metres, 
a circle is drawn. At what points will it cut the other side? 

13. Given an angle of 55°, and a point on one of its sides 
distant 171.6 from the vertex. A circle with its centre at this 
point is required to cut out a length of 85 from the other side 
of the angle. What must be the radius? 

, 14. A base-line 250 metres long is viewed from a point 100 
metres from one end and 200 metres from the ofher end. 
What angle will it subtend? 

15. The sides of a triangle are respectively 9, 11 and 13. 
Find its altitude above the longest side as a base. 

16. A quadrilateral whose sides are AB, BG, CD and 
DA is inscribed in a circle. Show that the angle A is given 
by the equation 

_ AB' - BC + DA' - CD' 
^^^ ~ 2{AB . AD-\-BG.CD) ' 
and find the corresponding expressions for the other angles. 

17. From the preceding expression show that the angle 
A may be found from the equation 

tan'-.l- (^--^-^^(^-^^) 
2 ~ {S - BC)(S - CDY 

where S represents half the perimeter of the quadrilateral. 

18. An inscribed quadrilateral ABCD has side AB = AD 
and B0= CD. Show that 

(Diagonal A Cf = AB' + BC\ 

19. A quadrilateral ABCD has 

Angle A = 49° 36'; Angle B = 86° 5'; 

Angle C = 120° 48'; Angle D = 103° 31'; 

Side AB = 40.29 ; Side CD = 25.32. 
Find the other two sides. 

Begin by producing the unknown sides AD and BC until they meet, 
thus forming two triangles. 

20 A cow is tethered by a rope 55 feet long to the vertex 
^ of a triangular field whose sides are: AB = 60 teet, 



SOLUTION OF TRIANGLES IN GENERAL. 65 

^C' = 75 feet, BC = ^0 feet. Over how many square feet 
can she graze? 

21. A tower 108 feet high stands at the vertex ^ of a tri- 
angular field whose sides are: AB = 255 feet, AC = 312 feet, 
and BC = 239 feet. What angle does the side ^C subtend 
as seen from the top of the tower? 

22. Near the foot of a flag-staff 96.3 feet high are two 
posts, the one, A, 82 feet toward the north, the other, B, 
155.6 feet toward the west. What is the shortest distance 
from the top of the flag-staff to the line AB? 

23. Two flag-staffs are 203 feet apart. From the middle 
of the line joining them the altitude of the higher is double 
that of the lower, but on going from the middle point 43| 
feet toward the lower the altitudes are equal. What is the 
height of each? 

24. Two light-houses stand near each other on a plane. 
Their altitudes, each measured from the base of the other, 
are 46° 6' and 33° 45' respectively, and the straight line be- 
tween the summits measures 87 feet. What is the height of 
each, and their distance apart? 

25. Two halyards of a flag-staff are respectively 228 and 
196 feet long, and the longer meets the ground 40 feet farther 
away than the shorter does. What is the height of the staff? 

26. A triangular field bounded by a road was defined as fol- 
lows: the second side made an angle of 41° with the road 
and was 363.8 yards long, and the third side ran 275 yards to 
the road. At what two points might it have met the road? 

27. Prove that in any triangle ABG 

tan IC = ^"|"^7 - cot iB. 

28. In a triangle is given: the base = 9.622, an angle at 
one end of the base 18° 47', and the sum of the two 
15.023. Find the parts of the triangle. 

If a triangle is right-angled at C, prove the equations: 

29. cos' \a = ^^. 31. cos 2ar = f' " ^\ , 

30. cos (or -y5) = ^ . 32. tan (a - p) ^' ~ ^' 



2a& 



66 PLANE TRIGONOMETRY. 

63. Case IV. Given two sides and the in- 
eluded angle. 

Theorem IV. A s the sum of any two sides 
is to their difference, 
so is the tangent of half the sum of the a7igles opposite these sides 
to the tangent of half their difference. 

Proof From the equation 

b : c :: sin J3 : sin y, (§ 58) 

we have, by composition and division, 

i -\- c : I? — c :: sin ^ -\- sm y: sin y5 — sin ^. 
But sin p -\- sin y := 2 sin l{/3 + y) cos ^{/3 — y); (§ 55) 
sin /3 — sin y = 2 cos ^{/3 -}- y) sin |^(y5 — y). 

Substituting these values, and expressing the proportion as a 

fraction, 

i -{- c _ sin |(^ + y) cos i{/3 — y) ^ 

i-o .., _ 

(§25,3,5) 



cos i(/S + r) sin i(/J - 
tani(/J+K)coti(^- 
tanf(/?+ r) 


-r) 
-y) 



tan ^{j3 — y) 
Therefore 
Z> + c : 5 - c : : tan i(/5 + :r) : tan i(y5 - y). Q.E.D. (16) 
The solution is now obtained as follows: We have 
L(/? -^y) = 90° - la; 
. tan i(/? + r) = cot 1^. (§ 57) 

Because the angle a is given, the only unknown term of 
the proportion is tan |-(y5 — y). This is given by the equa- 
tion 

tan i(^ - k) = ^ tan i(^ + ;.), (17) 

which is derived from the proportion (16). 

By this equation we obtain ■!■(/? — y), which being added 
to and subtracted from ^(/? + y) gives ft and y. The remain- 
ing side of the triangle may then be found by Case I. 

Example. Given h = 4.567, c = 3.456, a = 56° 7'.8; 
find the remaining parts. 



SOLUTION OF TRIANGLES IN GENERAL. 67 



h, 4.567 


180° 


c, 3.456 


a, 56° 7'. 8 


h - c, 1.111 


{P + ;/), 123° 52'.2 


h -\- c, 8.023 


i(/? + ;/), 61°56M 


log (J - c), 0.0457 


i(/?- r), 14° 33'. 6 


co-log {h + c), 9.0957 


ft, 76° 29'. 7 


log tan iCft + ;/), 0.2731 


;/, 47° 22'. 5 


logtaiii(/J- y), 9.4145 


\ogc, 0.5386 


log J, 0.6596 


log sin ;/, 9.8668 


log sin A 9.9878 


log Mod., 0.6718 


log Mod., 0.6718 
a = 3.899 


log sin a, 9.9192 


log a, 0.5910 



EXERCISES. 

1. Gwen ^=12.34, 5=43.21, ;k=34° 12'; find rem. anglea 

2. " 5= '/S, c= 4/3, «=35° 53'; 

3. '' 6^=35.79, c=1.2468,/5=:97° 53'; 

4. '' ^.=189, Z'=114.75,:r=107° 48'; " 

5. '' 5= 2956.2, c = 9090.8, a = 98° 29'.6; find ft, y 
and c^. 

6. A surveyor lays off two lines from the same point: the 
one due north, 279.25 metres, the other east 15° north, 
109.262 metres. How far apart are the ends of the lines, 
and what is the direction of the line joining them? 

7. The sides of a parallelogram are 26 and 15, and one 
angle is 126° 52'. 2. Find the lengths of the two diagonals 
and the angles which they make with the sides. 

8. Given the two diagonals d and c/' of a parallelogram 
and the angle s which they form; express the sides and angles 
of the parallelogram algebraically and compute them for the 
special case ^Z = 5, ^' = 6, f «= 49° 18'. 

9. The two parallel bases of a trapezoid are 42 and 32 re- 
spectively ; one oblique side is 15, and it makes an angle of 
65° with the longer base. What angle does the other oblique 
side make with the longer base ? 



68 PLANE TBIOONOMETBT, 

10. Two sides of a triangle are respectively AB = 27.52, 
AC = 32.59, and the medial line from B to the middle of 
AC = 26.20. Find the three angles of the triangle. 

11. The two parallel sides of a trapezoid are 42 and 34, 
and the non-parallel sides are 22 and 26. Find the angles 
and the diagonals. 

12. Within a field 125 metres square is a post distant 49 
metres from one corner and 115 metres from the adjacent 
corner. Find its distance from the other two corners. 

13. In the same field is a second post distant 67 metres 
from one corner and 122 metres from the opposite corner. 
Find its distance from the other two corners . 

14. Two ships sail from the same port, the one going 
IST.E. at the rate of 12 knots an hour, the other E. 12° S. at 
the rate of 9 knots an hour. How far apart, and in what 
direction from each other, are they at the end of 24 hours? 

15. In a quadrilateral A BCD are given: 

Side AB = 66.19; Side BG = 47.92; 

Side CD = 43.07; Side DA = 80.04; 

Angle A = 57° 22'. 

Find the three remaining angles, the angles B and D each 
having two values. 

16. Two sides of a triangle are in the ratio m : n. Express 
the ratio of the tangent of half the sum of the opposite angles 
to the tangent of half their difference. 

17. Two sides of a triangle are in the ratio 3 : 7 and form 
an angle of 36° 37'. Find the other two angles. 

18. The lines from a point within a triangle to its three 
vertices form three equal angles with each other, and their 
lengths are p, q and r respectively. Express the squares of 
the sides of the triangle in terms of p, q and r. 

19. Two adjacent sides of a parallelogram are a and h, and 
they form an angle q). Exprej|3 the lengths of the diagonals 
in terms of a, h and cp. 

20. The two diagonals of a quadrilateral ABCD form an 
angle A OB — 72° and intersect in a point 0, which divides 
them into the segments Oyl =3.298; 0(7=4.120; 0^=4.007; 
QD — 2.924. Find the four angles of the quadrilateral. 




AREAS. 69 

Areas of Triangles and Polygons. 

64. Theoeem. The area of a triangle is equal to half 
the product of any ttuo sides hy the sine of their included angle. 

Proof. It is shown in geom- 
etry that the area is half the 
base into the altitude. Now in b^ 

the figure, 

Altitude ^ = & sin y. 
Therefore, a being the base, ^ « 

Area = \ah — \dh sin y. Q.E.D. 

Cor. 1. If two triangles have two sides of the one respec- 
tively equal to ttuo sides of the other, and the angles which 
these sides for7n supplementary, the triangles will be equal in 
area. 

For the sines of the supplementary angles are equal (§ 38). 

Cor. 2. Since we may take any one side- as a base, if we 
call h, h', h" the altitudes aboye the respective sides a, i 
and c, we shall haye 

ah = W = ch'', 
and ah sin y = be sin a — ca sin /?. 

For these expressions are each double the area of the tri- 
angle. 

EXERCISES. 

1. Given a = '75, b = 29, ^ = 16° 15'. 6; find the remain- 
ing parts and the areas of the two triangles which may be 
formed. 

2. Express the area of a parallelogi'am of which two ad- 
joining sides are of given length, a and b, and make with each 
other a given angle S. 

3. In a parallelogram is given a diagonal of length d, and 
the angles 6 and cp which the diagonal makes with the two 
sides adjoining it. Express the area of the parallelogram. 

4. Express the area of a parallelogram in terms of the 
lengths d and d' of its diagonals, and the angle s at which 
they intersect. 

5. In a quadrilateral are given the four sides, a = 25.63, 
b = 24.09, c = 9.92, d = 29.97, and the angle, 78° 25', which 



70 PLANE TRIGONOMETRY. 

the sides a and h form, with each other. Compute the angles 
and the area of the quadrilateral. 

6. Show that in a right triangle the altitude above the 
hypothenuse is represented by 

■Jc sin 2 a: or -Jc sin 2/?, 
a and /? being the oblique angles, and c the hypothenuse. 

7. Show that the three altitudes of any triangle may be 

represented by the equations 

^ sin 6 sin y 
h —a -. ^; 

sm a ' 

, - ^ sin y sin a 

sm ' 
^// ^ ^ sin a sin /? ^ 

sin y 

8. Express the area of a triangle in terms of a base, c, and 
the two adjacent angles, a and ^. . ^ sin a sin ft 

AnS. -i-C —. 7 ; 77,, 

^ sm (a: + /J) 

9. A farmer had a rectangular field 300 X 250 yards, 
which he had to divide into two equal parts by a fence from 
a point 55 yards distant from one corner and on the longer 
side. What will be the length of the fence and the angle 
it makes with the longer sides? 

10. A ship heading due north and sailing through the 
water at the rate of 5 knots an hour was really in a current 
moving northeast at the rate of 3 knots an hour. What was 
the actual direction and speed of the ship? 

. 11. Compute the area of a regular pentagon each of whose 
sides is 16 feet. 

12. Three circles whose radii are 4, 5 and 6 feet respec- 
tively, mutually touch each other externally. What is the 
area of the space enclosed centrally between them? 

13. What is the side of that square whose area is equal to 
the area of a regular octagon with sides 12 feet long? 

14. A circular field 550 feet in diameter is divided into 
two parts by a straight fence passing 55 feet from the centre. 
What is the area of each part? 

15. A cow is tied by a rope 90 feet long to one corner of 
an oblong field 80 feet wide, but longer than the rope. Over 
how many square feet can she graze? 



AREAS. 71 

16. The home of a man lost in the woods was 5000 feet 
due north from where he stood. He wandered southeast 6500 
feet. What was then his distance and direction from his 
home? 

17. A triangle of which the base is 825 feet and the sides 
are respectively 758 and 696 feet is to be diyided into two 
equal parts by a line from the vertex to the base. What is 
the length of the line, and to what point of the base must it 
be drawn? 

18. The same triangle as in the preceding example is to be 
divided into two parts in the ratio of 3 to 2 and in the same 
way. What must be the length of the line, and where may it 
terminate? 

Note, There may be two answers. 

19. A farmer had a triangular field of which the sides were 
100, 120 and 136 feet. He tied a cow to each corner by a 
rope of such length that each cow could just meet the other 
two. What was the length of each of the three ropes, and 
over how many square feet could each cow graze? 

20. The angular elevation of a monument above a plane, 
as seen from a point x due south of it, is 30°. Going due 
west from x 120 yards the elevation is 22°. What is the 
height of the monument? 

21. At a distance of 80 yards from the foot of a tower a 

mark 50 feet high on its side has an angular elevation half 
that of the tower. What is the height of the latter? 
-^ 22. Two towers 50 yards apart stand on a horizontal 
plane. A person in a line with them and 100 yards from the 
nearer finds their apparent altitudes to be the same. Walk- 
ing toward them 80 yards, the apparent altitude of the 
nearer is double that of the farther. What are the heights 
of the towers ? 

23. Find the sides of that right triangle whose sides are 
in arithmetical progression, and whose perimeter is 72 feet. 

24. Find the angles of that right triangle whose sides 
are in geometrical progression. Begin by showing that in 
such a triangle the sine of one oblique angle is equal to the 
tangent of the other. 



CHAPTER VII. 

PROBLEMS IN TRIGONOMETRIC COMPUTATION AND 
ANALYSIS. 



65. Problem I. Having two equations of the form 
r sm cp = a, 
r cos (p = i, 

where a and h are given numbers, it is required to compute r 

and cp. 

Solution, Dividing the first equation by the second, we 

find 

sin fl) a 

— = tan CD z=z —, 

cos<p ^ 

From this value of tan q) we find ^ itself, then the sine or 
cosine of % and then r from either of the equations 

_ a __ 5 
"" sin cp ~ cos (p' 

Example. If r mi cp — 332.76, and r cos 9? = 290.08, 
it is required to find r and cp. 

log sin cp, 9.8772 (5) log cos cp, 9.8176 (6) 

log r sin cp, 2.5221 (1) log tan cp, 0.0596 (3) 

log r cos cp, 2. 4625 (2) cp, 48° 55' (4) 

log r = log r sin q) — log sin cp, 2.6449 (7) 

log r = \ogr cos cp — log cos cp, 2.6449 (8) 

r, 441.5 (9) 

The numbers in brackets show the order in which the numbers of 
the computation are written. In writing log r sin cp and log r cos cp 
spaces are left for inserting log sin cp and log cos cp after q) is found, so 
that either of the latter may be subtracted to obtain log r. It is gener- 



PROBLEMS IN TRIGONOMETBIG COMPUTATION. 73 

ally best to obtain r from both r sin <p and r cos <p, because then if any 
mistake is made in (p it will be shown by a difference in the results. 

On the other hand, a practical computer will not write down either 
sin q) or cos q), but will subtract them in his head and write down log r 
only. 

EXERCISES. 

Find r and q) from the following equations: 

1. r sin <7? = 1.297 43 and r cos ^ = 6.002 4. 

2. sin = 0.082 19 and cos = 0.128 8. 

3. sin =194 683 and cos =8460.7. 

66. Distinction of Qicadrants. In tlie preceding ex- 
amples we haye supposed r sin cp and r cos (p to be positive, 
and have taken cp in the first quadrant. But either or both 
of these quantities may be negative. Whatever their signs 
there are always two values of q?, differing by 180°, corre- 
sponding to any given value of tan (p (§ 41). Hence the pro- 
blem admits of two solutions in all cases. In the one r will 
be positive, in the other negative. 

But in practice only the positive value of r is sought. 
This being the case, sin q) and cos cp must have the same 
algebraic signs as the given quantities r sin q) and r cos cp re- 
spectively. Now consider each case in order: 

I. r sin (p and r cos cp doth positive. The angle (p must 
then be taken in the first quadrant, because only in this quad- 
rant are sin cp and cos cp loth j^ositive. 

II. r sin q? positive and r cos q) negative. Sin q) is posi- 
tive only in quadrants (1) and (2) (§ 38), and cos q) is nega- 
tive only in quadrants (2) and (3). Hence the requirement 
of signs can be fulfilled only in the second quadrant, and 

90° < q) < 180°. 

HI. r sin q? and r cos q) loth negative. The only quad- 
rant in which sine and cosine are both negative is the third. 
Hence in this case 

180° < q) < 270°. 

IV. r sin q) negative and r cos q? positive. The only quad- 
rant in which sin q? is negative and cos q> positive is the 
fourth. Hence in this case 

270° < q}< 360°. 



74 PLANE mmONOMETUT. 



EXERCISES. 


Find r and q) from the following equations: 


1. rsin ^ = — 237.09 and 


r cos 9? = + 192.91. 


2. '' -+2713.8 and 


- 9269.2. 


3. '' - 1.9634 and 


" - 0.09654. 


4. " - 3.6925 and 


" + 396.72. 


5. " - 18.005 and 


- 2.6943. 


6. Given r sin (^ + 47° 


50') = 7.2693 


L T cos {cp + 47° 


50') = - 12.2916, 



and 

to find r and cp. 

Note. In this last exercise compute the value of (9 -\- 47° 50') as if 
it were one quantity, and subtract the angle 47° 50' from the result. 

7. Given r sin (0 + ic) = - 249.88, 

rcos \e + a;) = - 92.62, 

u sin {d -x) = 702.02, 
and u cos (6 — x) =^ 516.93, 

to find the values of r, u, 6 and x. 

8. Given r (sin 6 + cos 6) = 298.07 
and r (sin 6 - cos 6) = 96.04, 
to find r and 6. 

61. Problem II. Jb reduce an expression oftlieform 
a^in 6 -\- h cos 6 (1) 

^ « monomialy a and l leing given quantities. 

Solution. Determine the values of two auxiliary quantities 
h and £ from the equations 

^ cos £ = a, 
h sin £ = 5, 
as in Problem I. 

The given expression will then become by substitution 
h cos £ sin + ^ sin £ cos = ^ sin (6 + £). 

We might equally well have supposed h ^\x\. b = a and 
^ cos £ = 5, when the expression (1) would have become 
h cos {Q — £). 

Example. Reduce the expression 

1239.3 ^\\ix — 724.6 cos aj 
to a monomial. 



PROBLEMS IN TRIOONOMETRIG COMPUTATION. 75 



^ sin £ = 


- 724.6 


log = - 2.8601 


^ cos £ = 


1239.3 


log = 3.0932 




log tan a,- 9.7669 






£,- 30° 19' 






log cos f, 9.9362 






log/t, 3.1570 






h, 1435.5 



We therefore have 

1239.3 sin x - 724.6 cos x = 1435.6 sin {x - 30° 18'. 8). 

EXERCISES. 

Reduce to monomials the expressions: 

1. 27.615 cos /< — 23.208 sin /i. 

2. 3. 600 3 sin {d - T 52'. 6) + 5.907 sin (6 + 53° 57'. 6). 

cos ^ sin 6 
3.^-- + -^. 

4. \ sin (6 + 5°) - ^ (9 - 5°). 

Find X from the equations: 

5. 8 cos ic + 3 sin a; = 7. 
sin a; cos a: _ 1 

' r25 ~ 2.50 ~ 2 ' 

Begin by reducing the first member to a monomial, thus reducing 
the equation to the form a sin or cos {x -{- ?i) = b. Then we find 

{x -\-h) = -, from which x-\- his found by the tables. 

cos Gi 

68. Peoblem III. From the equations 

r cos J3 cos X = a, ^ 

r cos /? sin A = b, > (1) 

r sin /J = c, ) 

to find the values of r, (3 and X, the values of a, l and c being 
given. 

Metliod of Solution, Dividing the second equation by the 
first we obtain 

tan A = -; (2) 

a 



76 PLANE TBIGONOMETRT, 

from this equation we find X, and then sin A or cos \ from 
the tables. Then 

r cos /? = T- = ~. — r- (3) 

cos A sm A ^ ' 

can be computed. The value of r sin /? being given by the 
third equation (1), the values of r and fi are found from (3) 
by Problem I. 

Example, Find r, fi and A from the equations 
r cos /? cos A = — 53.953; 
r cos /? sin A = + 197.207; 
r sin /5 = - 39.062. 

log r cos /5 cos A, — 1.7320 log cos fi, 9.9922 

log r cos /i? sin A, 2.2949 log r cos y^, 2.3106 

log sin A, 9.9843 log r sin /?, - 1.5918 

log tan A, — 0.5629 log tan /?, — 9.2812 

A, 105° 18'.0 log r, 2.3184 

log r cos A 2.3106 r, 208.15 

A - 10° 49' 

EXERCISES. 

Find the values of r, p and A from: 

1. r cos /J cos A = 1.2718; 
rcos/? sin A = — 0.9815; 
rsiny^ = 0.8900. 

2. r sin/? sin A = 19.765; 
r sin y^ cos A = — 7.192. 
r cos /? = 12.124. 



PAR.T II. 

SPHERICAL TRIGONOMETRY. 



CHAPTER I. 

FUNDAMENTAL RELATIONS. 



69. Def. Spherical trigonometry treats of the relations 
among the six parts of a trihedral angle. 

Def. The six parts of a trihedral angle are its three face- 
angles and its three edge-angles. 

70. Representation of a trihedral angle ly a spherical tri- 
angle. If be the vertex of a trihedral angle, and OM, OP 
and OQ its three edges, we may 
construct a sphere having its 
centre at 0, and having an 
arbitrary unit-radius OA. The 
spherical surface will then cut 
the edges at the three points A, 
B and C, equally distant from 
0. 

The three faces Oi/P, OP $, M- 
O^if will intersect the spherical surface in three arcs of great 
circles, AB, BG, OA, which arcs form a spherical triangle. 

By geometry the three angles A, B and (7 of the spherical 
triangle are equal to the respective edge-angles formed along 
OM, OP and OQ oi the trihedral angle. Also, the arcs AB, 
BO and OA, which form the sides of the triangle, measure 
the respective face-angles MOP, POQ, QOMoi the trihedral 
angle. 

Therefore the six parts of the trihedral angle are repre- 
sented by the corresponding parts of the spherical triangle, and 
the relations among the parts of the one are the same as the 
relations among the parts of the other. 




78 SPHERICAL TBIQONOMETRY. 

The term spherical trigonometry is used because the in- 
yestigation of a trihedral angle is generally made by means of 
the spherical triangle. 

71. Eacli side of a spherical triangle is supposed less than 
180°, unless otherwise expressed. 

KoTE 1. A trihedral angle, with its corresponding spherical triangle, 
may be readily constructed as follows: Cut a circular disk of pasteboard 
or stiff paper, from four to six inches or more in diameter. From this 
disk cut out a sector of any magnitude. It will be well to have several 
disks with sectors ranging from 45° to 200° cut out. Divide the re- 
mainder of the disk by two radii into three sectors, such that the great- 
est shall be less than the sum of the other two. Bend the disk along 
each of the two dividing radii, cutting the latter part of the way through 
if necessary, and bring the extreme radii together. We shall then have 
a figure like 0-ABG of the preceding diagram, the three plane sides form- 
ing the trihedral angle, and the three arcs bounding the edge of the disk 
forming the spherical triangle. 

Note 2. The student should guard himself against considering any 
figure of which any side is a small circle of the sphere as a spherical 
triangle. For example, the figure formed by two arcs of meridians and 
a parallel of latitude is not a spherical triangle. Such figures do not 
represent the parts of a trihedral angle, and so do not correspond to the 
definition of a spherical triangle. All the important problems con- 
nected with them may be reduced to problems of spherical trigonome- 
try, so that there is no need of giving them special consideration. 

EXERCISES. 

The following exercises are introduced to test the student's funda- 
mental conceptions of spherical geometry, and especially of the relations 
of great circles of the sphere. Their successful performance will show 
that he is prepared to take up the subject of spherical trigonometry with 
advantage. A globe, on which figures may be drawn at pleasure, will 
be of great service in assisting his conceptions, and should be made use 
of as far as practicable. 

1. A and A' are two opposite points on a sphere. If any 
third point X be taken on the sphere, to what constant arc 
will the sum XA -f- XA' be equal, and what will be the angle 
AXA'? 

Note. Opposite points are those at the ends of a diameter. 

2. If one side of a spherical triangle be equal to a semi- 
circle, what relations will subsist between the other two sides? 
What will be the magnitude of the opposite angle? 

3. Let A, B and G be the three vertices of a spherical tri- 



FUNDAMENTAL RELATIONS. 



79 




' V 




The eight spherical tri- 
angles formed by three 


great circles. 





angle; a, h and c, the sides opposite these yertices respective- 
ly; A', B' and G\ the points opposite the vertices. It is then 
required — 

(a) To show that when each side of the original triangle 
ABC'\^ produced into a great circle^ eight 
triangles will be formed. 

(^) To express the sides of each of 
these eight triangles in terms of a, l and 
c, making use of the theorem that any 
two great circles intersect each other in 
two opposite points. 

(c) To express the angles in terms of 
the angles of the original triangle, which 
we may represent by the letters A, B and 
G marking their vertices. 

{d) It being found that the eight triangles are divisible 
into four pairs, such that the sides and angles of each pair are 
equal, it is required to show the relations of each pair. 

4. If one angle of a spherical triangle is A, show that the 
sum of the other two angles is contained between the limits 
180° - A and 180° + A. 

Note, If the student finds any difficulty in this question, he may- 
begin by supposing tlie triangle to be isosceles, and the two equal sides 
to increase from 0° to 180°. 

5. If the three sides of an equilateral spherical triangle be 
continually increased, what is the limit of their sum? What 
is the limit of the angles as the sum approaches its limit? 

•72. Fundamental Equations. Let us put : 
«, h, c, the three face-angles of the trihedral angle — that is, the 
angles subtended by the three 
sides of the spherical triangle; 
A, B, (7, the opposite edge-angles 
of the trihedral angle, or the 
angles of the spherical triangle. O, 
Then, if 0-ABG be any trihe- 
dral angle, we have 

a = angle BOG; 
h = angle GO A; 
c = angle A OB. 




80 SPHERICAL TBiaONOMETBY. 

Through any point A of OA pass a plane perpendicular to 
OA, and let B and C be the points in which it meets the 
other two edges. We shall then have 

A — angle BA G, 
while OAB and O^C will both be right angles. (G-eom.) 
In the triangle BOG we have 

BG'= 0B'+ 0G'-20B.0G cos a. (§60) 

In the triangle BA G we have 

BG' = AB' + AC' - 2AB.AG X cos A. 
Equating the above two values of BG', and transposing, we 
find 

20B . OG cos a = 0B'-AB'-{-0G'-'A G' + 2AB .AC cos A. 

But, because OAB and OA G are right angles, 

OB' - AB' = OA'; 

OG'-AG'= OA', 

Substituting these values, and dividing by 20B . OG, we have 



cos a = 

Now 


OA OA AB 
' OB' OG ' OB 

OA 

^^ = cos.; 

OA 

^^ = cos&; 

AB 


, ^^cos^, 


Therefore 


AG . ^ 





cos a = cos 5 COS c 4- sin ^ sin c cos -4. (a) 

By treating the other edges in order in the same way, we 
obtain two more equations, which may be written by simply 
permuting a, h and c, and A, B and C — that is, by substitut- 
ing for each letter the one next in order, a following c and A 
following G. Thus we have the system of three equations: 
cos a = cos i cos c -{- sin d sin c cos A ; j 
cos b = cos c cos a -\- sin c sin a cos B; > (1) 

cos c = cos a cosh -\- sin a sin 5 cos C. ) 
These three equations are the fundamental equations of 
spherical trigonometry, because by means of them, when three 



FUNDAMENTAL BELATI0N8. 81 

parts are given, the other three may be found. For practical 
application they are transformed and simplified in numerous 
ways, as will be shown subsequently. 

73. Theorem of Sines. In a spherical triangle the 
sines of the sides are proportional to the sines of the opposite 
angles. 

Proof. Let 0-ABC be the trihedral angle of the spheri- 
cal triangle, and let A be any A-^-^" 
point on the edge OA. ^^^^/\\ 

Through A pass a plane ^^^^ / \\ 

perpendicular to the edge OB, ^^^ I \ *^ 

intersecting ths faces AOB q^P / i \^ 

and BOC in the lines AB and ^^' / — \~j-~ 

BP. \^ / ;/ 

Through A pass another ^-vA--- " ? 

plane perpendicular to OC, in- ^\ 

tersecting the faces AOG and COB in the lines ^(7 and CP. 
AP will then be the line of intersection of these planes. 
Because the planes ABP and A OP are respectively per- 
pendicular to the lines OB and 00, they are each perpendicu- 
lar to the plane BOO of these lines (Geom.). Therefore their 
line of intersection, AP, is also perpendicular to this plane, 
and the triangles APB and APO are right-angled at P. 
Hence 

AP = AB sin ABP = AB sin B, 
Also, because ^^ is perpendicular to OB, 

AB = OA sin BOA ~ OA sin c. 
Therefore AP = OA sin c sin B. 
We find in the same way 

AP = OA sin d sin 0; 
whence 

sin c sin ^ = sin b sin O, 
or 

sin c _ sin 5 
sin ~ sin B' 
We may show in the same way, by permuting the parts, 

sin a _ sin b _ sm c .^. 

sin A ~ sinB ^ sin C ^ ' 



82 SPHEEIGAL TBIGONOMETBT. 

These equations may be expressed in the form of a multiple 
proportion: 

sin a : sin 5 : sin c = sin ^ : sin ^ : sin G 

Polar Triangles. 

74. Def, When two triangles are so related that the 
vertices of the one are the poles of the sides of the other, the 
one is said to be the polar triangle of the other. 

It is shown in geometry that the relation of a triangle to 
its polar triangle is reciprocal; that is, if X and P" are two 
triangles, and Y is the polar triangle of X, then X is the 
polar triangle of Y. This reciprocity arises from the theorem: 

If A, B and C he the three poles of the sides QR, RP and 
PQ of a triangle PQR, then P, Q and R ivill he the poles of 
the sides BC, CA and AB of the triangle ABC. 

This theorem is readily proved by the geometry of the 
sphere. 

Since every great circle has two poles, one at each end of 
a diameter, it follows that the three sides of a triangle have 
six poles in all. We may form a polar triangle to ^^C by 
taking either of the poles of AB, either of the poles of BC, 
and either of the poles of CA, and joining them by arcs of 
great circles. Hence there are eight possible polar triangles 
to every given triangle. To avoid c' 

doubt which triangle is to be / ^'--. 

chosen, we take for each vertex /'' ^ ^, 

of the polar triangle that pole of /' / \ ^\^ 

each side of the given triangle /' / \ ^,^ 
which is on the side toward the / / \ 

interior of the triangle. / / 3^ 

For example, if ABC is the A ^ ..----'B' 

given triangle, we take that pole 
oi AB which is on the side toward (7, and so with the other 



EXERCISES. 

1. What must be the sides and angles of a triangle that it 
may coincide with its polar triangle? 

2. Show that if each side of a triangle is greater than 90° 



FUNDAMENTAL RELATIONS. 83 

the polar triangle will fall wholly inside of it, and if each side 
is less than 90° it will be wholly within its polar triangle. 

3. If two sides exceed 90° and the third side is less than 
90°, what will be the character of the polar triangle, and how 
will it be situated relatively to the given one? 

75. Use of the Polar Triangle. By geometry each side of 
any triangle is the supplement of the opposite angle of its 
polar triangle, and vice versa. This principle is applied to 
find new relations between the parts of a triangle in the fol- 
lowing way: 

1. We imagine ourselves to construct the polar of the given 
triangle. 

2. We write any or all the equations between the parts of 
the polar triangle. 

3. We substitute in these equations the supplementary 
parts of the given triangle, and thus obtain equations between 
these parts. 

Let us put a', &', c', 

the sides and opposite angles of the polar triangle. Since the 
general equations (1) are true for every triangle, they are true 
of this polar triangle. Hence 

cos a' = cos l' cos c' + sin l' sin c' cos A\ 
But the polar triangle is so related to the original triangle 
that 



a' = 180° - A, 


A' = 180° - a; 


V = 180° - B, 


B' = 180° - i; 


c' = 180° - 0, 


C = 180° - c. 


Therefore 




cos a' =■ — cos A, 


cos A' = — cos 


cos y = — cos B, 


cos B* = — cos 


cos c' = — cos G, 


cos G' — — cos 


and sin a' = sm A, 


sin ^' = sin a\ 


etc. 


etc. 



Making these substitutions in the equations (1), we find 
cos ^ = — cos ^ cos G -\- sin B sin C'cos a\ 
cos ^ = — cos C cos ^ + sin G sin A cos J; \ (3) 
cos (7 = — cos J. cos ^ + sin J. sin B cos c. 



84 SPHERICAL TBIGONOMETRT. 

This process may be generalized thus: 

From every relation letween the parts of a spherical tri- 
angle we may derive another relation ly changing the cosine of 
each part into the negative of the cosine of the opposite part, 
and the sine of each part into the sine of the opposite part. 

But this relation will not always be different from the 
original one. If we apply the process to the equations (2), 
for instance, the same relations will be reproduced, each term 
being changed to its reciprocal. 

Transformation of the General Formulse, 

76. In the solution of spherical triangles we require for- 
mulae which shall express any three parts in terms of the re- 
maining three; the latter being supposed known, the former 
unknown. 

A set of such equations may be derived from the funda- 
mental equations (1) by substituting in any one the value of 
the cosine of a side obtained from another. Let us substitute 
in the third the value of cos a from the first. We shall have 

cos c = cos' bcosc -\- sin b cos d sin ccosA-\- sin a sin b cos C. 

Transposing cos c, noting that 1 — cos'^ b = sin' b, and divid- 
ing by sin b, we have 

= — sin ^ cos c + cos h sin ccosA-\- sin a cos C, 
which gives 

sin acos C = sin 5 cos c — cos b sin c cos A, (4) 

The theorem of sines (2) also gives 

sin asm G = sin c sin ^. (5) 

Comparing these two last equations with the first equation 
(1), we see that they form a set in which the second members 
contain only the parts b, c and A, that is, two sides and the 
included angle ; while the first members contain the third 
side, a, and one of the angles adjacent to it, namely, (7. 

•77. The equations (5), (4) and (1)^ therefore form a 
group. The following is their proper order: 
sin a sin = sine sin A ; 

sin « cos 0= sin bcosc — cos b sin c cos ^ [ (6) 
cos a = cosb cos c + sin b sin c cos 



A.) 



FUNDAMENTAL BELATIONS. 



85 



By using the angles b and B instead of c and C, we have, 
instead of (5) and (4), the equations 

sin a sin B = sin b sin. A; ) /^,n 

sin a cos ^ = sin <? cos b — cos c sin b cos A.)^^ 

By applying the same reasoning to the other sides and 
angles which we have applied to a, A, b, B and c, C, we find 



sin bsin C = sin c sin B; 
sin bcos C = cos csina — sin c cos a cos ^; 
sin i^ sin J = sin « sin ^; 
sin b cos ^ = sin c cos « — cos c sin a cos ^; 
cos b = cose cos 0^ -|- sin c sin a cos ^. 

sin c sin ^ = sin a sin C; 
sin c cos ^ = cos a sin 5 — sin a cos b cos C; 
sin c sin ^ = sin 5 sin C; 
sin c cos B = sina cos & — cos a sin 5 cos C; I 
cos c = cos acosb ~\- sin 6! sin b cos (7. J 



(7) 



(8) 



Then we obtain three similar sets through the interven- 
tion of the polar triangle. Applying the sets (6), (7) and (8) 
to the polar triangle, we find, by the rule of § 75, 



sin Asinb = sin B sin a; 
sin A cosb = cos B sinC -{- sin B cos C cos a; 
sin ^ sin c = sin (7 sin a; 
sin ^ cos c = sinB cos (7 + cos ^ sin (7 cos a; 
cosA= — cos ^ cos C+sin^ sin C'cos a, 

sin ^ sin c = sin sin b; 
sin B cose = cos C sin ^ -j- sin (7 cos ^ cos b; 
sin 5 sin cj = sin ^ sin 5; 
sin B cosa= sin Ccos ^ + cos (7 sin ^ cos 5; 
cos^= — cos (7cos^+sin C sin A cosb J 

sin Csin fi^ = sin A sin c; 
sin Ccosa = cos ^ sin 5 + sin ^ cos B cos c; 
sin Csinb = sin ^ sin c; 
sin Ccosb = sin ^ cos j5 + cos ^ sin B cos c; 
cos (7 = — cos J. cos ^-j-sin A sin 5 cos c. J 



(9) 



(10) 



(11) 



y (12) 



86 SPHERICAL TBIO0N0METB7. 

If in each of the sets of equations (6) to (8) we diyide the 
second equation by the first, and the fourth by the third, and 
clear of denominators, we shall haye the following additional 
equations: 

cot B sin A = cot ^ sin <? — cos c cos A: 
cot C sin A = cot c sin ^ — cos d cos A : 
cot C sin B = cot c sin a — cos a cos B: 
cot A sin B = cot a sin c — cos c cos B-. 
cot yl sin C = cot fl^ sin & — cos i cos C; 
cot ^ sin C = cot J sin r*^ — cos a cos 61 

If two sides and the included angle of a triangle are given, 
we may call these parts b, c and A, and the equations (6) and 
(6') will then enable us to compute a, B and C. 

If two angles and the included side are given, equations 
(9), (10) or (11) will enable us to find the other three parts. 
The mode of applying such equations will be explained in 
Chapter III. 

EXERCISES. 

1. Show that if we take the sum of the squares of equa- 
tions (6), botji members of the equation thus formed will re- 
duce identically to unity. 

2. What equations do we find by applying the theorem of 
sines to the polar triangle? 

3. If a = d = 0= 45°, 

show that tan A = tan B = 2 ~{- V2 ; 

cos c = i(l + Vi). 

4z. If two angles and the side between them are each 45°, 
find the other three parts. 

5. If one side is 60° and each of the adjacent angles 30°, 
find the remaining parts. 

6. If one side is 135° and the adjacent angles each 45°, 
find the remaining parts. 

7. When a = 30°, b = 60°, G = 45°, find the remain- 
ing parts. 



BIGHT AND QUADRANTAL TRIANGLES. 87 



CHAPTER II. 

RIGHT AND QUADRANTAL TRIANGLES. 



78. Fundamental Definitions. 

Def. A right spherical triangle is one which has a 
right angle. 

Def. A quadrantal spherical triangle is one which 
has a side equal to a quadrant. 

Def. A trirectangular triangle is one with three right 
angles. 

Def. A' birectangidar triangle is one with two right 
angles. 

Def A biquadrantal triangle is one with two sides 
each equal to a quadrant. 

79. Theoeem I. Every Urectangnlar triangle is also 
liquadrantal. --^ 

Proof Let J. 5 C be a spherical triangle 
in which angle B = angle = 90°. Then: 

Because angle ^ is a right angle, the 
pole of the great circle -SC is on the great 
circle BA; 

Because angle C is a right angle, this 
pole is on the great circle CA (Geom.). ^^^ 

Therefore the pole of BO is on both BA and 
therefore at their point of intersection A. 

Because A is the pole of BG, AB and AC ave quadrants. 

Q.E.D. 

Theorem II. Conversely, Every Mqnadrantal triangle is 
also Mrectangular. 

Proof Because every point of the polar circle of the 
point J. is a quadrant distant from A, and because AB and 
A C are quadrants, this polar circle must pass through both 
B and C. 




88 



SPHERICAL TRIGONOMETBT. 



But only ont great circle can pass through these points. 

Therefore BG i^ the polar circle of A, and A the pole of 
BG. 

Therefore the great circles AB and AG intersect BG at 
right angles. Q.E.D. 

Gor. Every triredangular triangle has three quadrants 
for its sides -y and, 

Conversely, Every triangle having three quadrants for its 
sides is triredangular. 

Theoeem III. In a hirect angular triangle the ollique angle 
is equal to its opposite side. 

Proof. Because the plane of the great circle BG intersects 
the planes of AB and of ^C at right angles, the arc BG 
measures the dihedral angle between the planes AB and BG. 

But the angle A is equal to this same dihedral angle. 

Therefore ^(7= angle A. 

Theorem IV. The polar triangle of a right triangle is a 
quadrantal triangle. 

This follows at once from the fact that the angles of the 
one triangle are the supplements of the sides of the other. 

Since in a right tri- 
M 




I 



80. FormulcB for Right Trian 
angle one of the parts, the right 
angle, is known in advance, if 
two other parts be given the re- 
maining three parts may be 
found. Only five parts have 
therefore to be considered. 

What we now want is to find 
equations between any three of 
these five parts, because by such an equation, when two of 
the parts are given, the third may be found. 

To find these equations let G be the right angle, and 
therefore c the hypothenuse. We seek for those equations, in 
the sets (6) to (12) of the last chapter, in which the angle G 
enters, and in which the equation contains only three other 
parts. We then suppose 

sin (7=1; 
cos C = 0; 
cot a = 0. 



BIGHT AND QUADBANTAL TBIANOLES. 



The set from which the required equations are taken, the 
number in the set, and the results, are shown as follows: 



From (6),, sin <2 = sin c sin A; (1) ) 
sin J = sin c sin B', (2) ) 
cos A = tan bcotc; (3) ) 
cos B = tan a cot c ; (4) f 
cos c = cos a cos b ; (5) 
cos 5 = cos ^ sin ^; (6)) 
cos-4 = coscf sin^; (7)) 
cos c = cot A cot B; (8) 
cot A = cot a sini ; (9) ) 
cot B = cot Z* sin a. (10) f 
These ten equations will be found to include all combina- 
tions of three out of the five parts a, h, c. A, B. From each 
of them we may determine any one part in terms of any two 
others: 



(6)., 

(6)„ 

(8)., 
(9),, 
(9)., 

(13)., 
(12)., 



for example, the first equation gives not only 
sin « = sin c sin A, 



but 



and 



sm c = 



sm a 
sin J.' 



sin^ 



sm a 



sm c 

Properly speaking, only six of these equations are really distinct, 
and the others can be derived from them by a mere interchange of letters 
between corresponding parts. For instance, since the same relation 
must hold between each oblique angle and its opposite side, the second 
equation may be derived from the first. 

The equations which are thus related are connected by braces in the 
formulae above. 

81. Napier^ s Rules. The six preceding formulae, which 
may be found difiScult to remember, have been included by 
Napier in two precepts of re- ^ 

markable simplicity, and easily 
remembered. 

Let us take for the five parts 
the sides a and l as before, and, 
instead of the other three parts, 
the complements of the oblique 
angles and of the hypothenuse. 




90 8PHEEICAL TBiaONOMETBT. 

The fact that the complements are understood is indicated 
by accenting the letters in the diagram. We suppose 

B' = 90° - B', c' = 90° - c; A' = 90° - a. 
Omitting the right angle, the five parts a, h, A', c', B' form a 
continuous series, B' being followed in regular order by a, 
Now if we select any three of these parts, one of two cases 
must occur. Either 

(1) The three parts all adjoin each other, as B\ a^h', a, h, 
A', etc., or 

(2) Two of the parts adjoin each other, and the third is 
separated from each of them by the remaining intervening 
parts. 

The middle part of the three in the first case, or the sepa- 
rated part in the second, is called the middle part 

In the first case the extreme parts of the three are called 
adjacent parts. 

In the second case the adjoining parts are called opposite 
parts. 

82. !N"apier's rules are: 

1. The sine of the middle part equals the product of the 
tangents of the adjacent parts. 

II. The sine of the middle part equals the product of the 
cosines of the opposite parts. 

The concurrence of the vowel a in tangent and adjacent, 
and of the vowel o in cosine and opposite, will help in remem- 
bering the relations. 

Examples. 1. Let the parts be the hypothenuse and the 
two adjacent angles, or c, A and B. 

The middle part is c', and A' and B' are adjacent parts. 

By the rule, 

sin (90° - C)- tan (90° - A) tan (90° - B), 
or cos c = cot A cot B, 

agreeing with the formula (8). 

2. Let the parts be a, A and G. The middle part is then 
a, and A' and C are the opposite parts. Therefore 

sin a = cos (90° - A) cos (90° - C) 
= sin A sin c, 
agreeing with the first formula. 



RIGHT AND QUADBANTAL TRIANGLES. 91 

3. Let the three parts be the two sides containing the right 
angle and one of the oblique angles, say a, h and A, Then h 
is the middle part, and the other two adjacent. Therefore 

'^ sin Z> = tan a tan (90° — A), 

or sin Icoi a = cot A, 

agreeing with the ninth formula. 

EXERCISES. 

1. Given A = 62° 29'.3, l = 25° 58'.8; find a. 

2. Given ^= 35° 29'.6, a= 75° 5'.3; find ^ and 5. 

3. Given^ = 43° 40'.5, c = 98° 29M; find^,^andj. 

4. Given a =148° 28'.2, A =101° 3'.9; find d and c. 

5. Given ^ = 50° 0'.8, ^ = 79°57'.3; find a, I and c. 

6. If, in a right triangle, we have given the sum of the 
sides a and 1) equal to s, derive the equation 

cos 5 + cos (s — 25) = 2 cos c. 

7. In a right triangle the hypothenuse c — 90°, and 
a-\- i = 120°. Find two values each of A, B, a and h, 

* In a triangle, right-angled at C, prove the relations: 

8. sin A sin 25 = sin c sin 2B. 

9. sin 2 A sin c = sin '^a sin B. 

10. sin 2^ sin 25 = 4 cos ^ cos B sin^ c. 

11. sin' \c = sin'^ -Ja cos' -J-J + sin' ^5 cos' ^a. 

12. sin (c — 5) = tan'i^ sin (5 + c). 

13. sin a oos d = tan ^^ sin (b + c). 

14. In a right triangle of which the oblique angles are 

A = 69° 23'.7, B = 60° 7'.6, 
find the length of the perpendicular from the right angle upon 
the base, and the angles which it forms with the sides. 

Using the results of §§ 27-29, express the cosines of the 
hypothenuse and of the sides when the oblique angles have 
the following values : 

B = 60°. 

B = 60°. 

B = 135°. 

B = 135°. 
and a = 60°, find cot A; cot B; cos c. 
and A = 60°, find the sines of b^ B and c. 



15. 


A = 60°; 


16. 


A = 45°; 


17. 


A = 45°; 


18. 


A = 135°; 


19. 


If b = 30° 


20, 


If « = 30° 



92 SPHERICAL TRIOONOMETRT, 

83. Isosceles Triangles. Any isosceles spherical triangle 
may be divided into two symmetrical right triangles by a 
perpendicular from its vertex upon its base. If we put 

c, each of the equal sides; 

G, each of the equal angles at the base; 

Z», the base, or third side; 

B, the angle at the vertex; 

P, the middle point of J; 

p, the length of the perpendicular BP from B upon h, — 
we shall have two right triangles in each of which tlie oblique 
angles are G and ^B, the hypothenuse is c, and the sides con- 
taining the right angle are p and ^l. The equations (1) to 
(10) of § 80 will then give 

sin ^l = sin c sin ^B', 

sin p = sin c sin C; 

etc. etc. 

EXERCISES. 

1. The equal sides of an isosceles triangle are each 45°, 
and the angle which they contain is 95°. Find the base and 
the angles at the base. 

2. If the base of an isosceles triangle is 95°, and the 
angles at the base each 45°, find the remaining parts. 

3. Prove that the angles at the base of an isosceles triangle 
are greater or less than 90° according as the equal sides are 
greater or less than 90°. 



SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 93 



CHAPTER III. 
SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 



84. The formulaB of Chapter I. suffice to determine any 
part of a spherical triangle when three other parts are given. 
But they are not in the form most conyenient for computation; 
we shall therefore show how they may be transformed to 
solve spherical triangles in any case. 

S^. Case I. Given the three sides. 

Required the three angles. 

From the first equation (1) of Chapter I. we obtain, by 

solving with respect to cos A, 

. cos fl^ — cos & cos c ,^. 

cos A = : — ^—. , (1) 

by which we can find A when a, h and c are given. 

To deduce a formula more convenient for computation 
add unity to each side of this equation. Then 

^ , . cos « — cos 5 COS c + sin 5 sin c 

1 + cos ^ = : 9 : ■ , 

sm c sm c 

„ - . , cos a — cos (J + 6) ,-. 

or 2 cos' ^A = ^1—^ — —^' (2) 

sm sm c ^ ' 

Subtracting the equation (1) from unity we find in the same 
way 

2 sin» iA = -cos^ + cos(&-.)^ 

sm sm c ^ ' 

Dividing (3) by (2), we have 

tan' iA = - co^ « + co« (^ - c) 

^ COS « — COS (5 + C) 

By § 55, Eq. (13), the terms of this fraction may take the 
forms 

— COS ^ + COS (5 — c) = 2 sin \{a + 5 — c) sin \(a — 5 + c); 
COS a — cos (Z> + c) = 2 sin \{a + 5 + c) sin J(— «+ l +c). 



94 



SPEEBICAL TRIGONOMETRY, 



Now, if we put 
we haye 



s = ^{a-\-'b-\- c), 



\(a-^l) - c) - s - c\ 
^{a — h -\- c) = s — 1}\ 

By substitution the value of tan' \A now becomes 

ten' ^A ^ '^ (^ - ^) f " '^ T 'l 
2 sij^ g sin (s — a) 

Also, from (3) and (2), 



sin i J. 



/ 



sin (s — &) sin {s — c) ^ 
sin h sin c 



cos 



1 J _ . / sin g sin (g — (x) 



(4) 



(5) 



2^ '^ sin & sin c 
Applying the same method to the other sides and angles in 
rotation, we shall obtain results which may be written by sub- 
stituting for each letter the next in order, as we have derived 
the equations (1) and (2), §§ 72, 73. Thus: 

c) sin (8 — a) 



tan' \B = '"^ 



tan' W = 



sin s sin (5 ~ i) 
sin (s — a) sin (s — d) 



(4') 



sin s sin (s — c) 
If we multiply both numerator and denominator of the 
second members of these equations by sin (s - a), sin (s - I) 
and sin (s — c) respectively, and if we put, for brevity, 
„ sin {s - a) sin {s - h) sin {s - c) 



we shall find 



sm s 



tan' ^A 



sin' (s 
then, by extr acting the square roots, 
p = l^csc 8 sin (s 



^)' 



a) sin (s - ^') sin (s - c); (6) 



tan J J. = 
tan i^ = 



tan ^G 



V 



sin (s — «y 
sin (5 — &)' 



(7) 



sin {8 — cY ) 
which is a convenient form for computation. 



SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 95 

Example of Form of Computation. Given the three sides 
« = 76° 29'.3, I = 93° 18'.6, c = 122° 7'. 7, 
to find the angles. 

a = 76° 29'. 3 s = 145° 57'. 8 log esc = 0. 2520 

Z>= 93 18.6 s-a= 69 28.5 log sin == 9.9715 

c = 122 7.7 5-^*= 5239.2. log sin = 9.9003 

, s-c= 23 50.1 log sin = 9.6065 



2s = 291° 55'. 6 



log y = 9.7303 

log^, 9.8652 logp, 9.8652 log p = 9.8652 
sm{s-a), 9.9715 sin {s-i), 9.9003 sin (5 - c), 9.6065 

tani^ 9.8937 tani^, 9.9649 tan -J6; 0.2587 
iA, 38°3'.5 -J^, 42°41'.2 iO, 61° 8'.3 

A, 76° 7' ^, 85° 22' C, 122° 17' 

As a check upon the accuracy of the computation we 

, , , , ,, T . sin ^ sin 5 , sin c 

should compute the yalues of -. 7, - — ^ and - — 7^ and see 

^ sm J. smB siti C 

if they agree. If they do not, there is some mistake in the 

work. 

sin a, 9.9878 sin d, 9.9993 sin c, 9.9278 

sin^, 9.9871 sin 5, 9.9986 sin (7, 9.9270 



0.0007 0.0007 0.0008 

The agreement is as good as could be expected. 



EXERCISES. 

Find the angles of triangles haying the following sides: 

1. a = 105° 6'.8; b = 93° 39'.9; c = 50° 20'.3. 

2. a = 122 27 .0; b = 104 25 .2; c = 125 .8. 
S.a= 9 62 .6; b = 15 8.8; c= 22 50.5. 
4:. a = 155 22; J = 75 13; c = 92 16. 

6. a= 23 37; 5 = 42 2; c = 58 19. 
6. If the sides of a spherical triangle are each 60°, show 
that the sine, cosine and tangent of the equal angles are: 

sine = —5—, cosine = ^, tangent = 2 V2. 
o o 



SPHERICAL TRIGONOMETRY. 



86. Case II. Given the three angles. 

Required the three sides. 
If we solve the first equation (3) of Chapter I. with re- 
spect to cos a, we find 

cos ^ + cos ^ cos G 

cos a = ^-^—^-- — 77 . 

sm B sm G 

Proceeding as in Case I.^ we find 

1 — cos fl^ , „ , — cos ^ — cos (5 + G) 

— = tan' ia = -:— )^ ^{ 

1 -\- COS a cos ^ + cos (B — G) 

- — ^ cos i{A-\- B -\- G) cos ^{-A -\. B -\- G) 

~ 2 cos ^\a-{- B - G) cos ^{a - B -\- G) ' 

Putting 

S=i{A + B-\- (7), 

and writing first the resulting equation and then the equations 
for the other sides, we have 



tan'^ ^a = — 



tan' ih = 



tan' ic 



cos iS'cos (S — A) 
cos (S - B) cos {8 - Gy 

cos S cos (8 — B) 
COB {8- G)cos{8- Ay 

cos 8 COS {8 — G) 



{a) 



cos (;S^ - A) COS {8 - BY J 

Multiplying both numerator and denominator of these 
equations, the first by cos (8— A), the second by cos (8 ■— B), 
and the third by cos {8 — G), and putting, for brevity^ 

p2 — cos 8 

~ cos (8 - A) cos (8 - B) cos (8 - GY 

the first equation will become 

tan' ia = P' cos'(/S -A), 
with corresponding equations for h and c. Hence, if we com- 
pute 

/ — nria Si 

:, (W) 



/ 



we shall have 



cos {8 - A) cos {8 - B) cos {8 - GY 



tan ^a= P cos {8 — A); 
tan-iJ = Pcos(8 - B); 
tan ^c = P cos (8 — G). 



(11) 



SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 97 



A = 76° 6'.7; 


B = 85° 22'.0; 


(7 = 


122°16'.6. 


A = 46 59 .3; 


B = 122 32 .6; 


(7 = 


139 .3. 


^ = 78 40.7; 


B = 102 29 .5; 


C = 


86 49.4. 


A = 158 30 .0; 


B = 58 14 .5; 


C = 


61 29 .1. 


A = dH 7.0; 


5 = 84 13 .6; 


C = 


75 18.8. 


A = 22 34 .2; 


5 = 39 54.2; 


C = 


125 23 .0. 



EXERCISES. 

Find the sides of triangles having the following angles : 

1. 

2. 

3. 

4. 

5. 

6. 

7. Show what the equations (a) become in the case of a 
trirectangular triangle, and thus show that such a triangle is 
also triquadrantal. 

8. Show from the same equations that in a birectangular 
triangle the oblique angle is equal to its opposite side. 

9. If the angles of a triangle are 

A = 120°, B = C = 60°, 
show from the equations (a) that the sides are given by the 
equations 

tana = — 2 V2; tan 5 = tan c = 2 V2. 

10. From the same equations find the sides of a triangle 
each of whose angles is 120°. 

11. From such of the preceding equations as you deem 
most suitable find the cosines of a, 5 and c when 

A = 60°; B = 120°; C= 135°. 

8*7. Case III. Given two sides and the included angle. 

Required the remaining parts. 
Let a and h be the given sides, and G the given angle. 
The following theorem affords the most convenient method 
when all three of the remaining parts are required. 

Theorem. In any spherical triangle we have the equa- 

tions 

sin ^c sin ^{A — B) = cos -J-C sin ^{a — i): 
sin ^c cos i{A — B) = sin ^C sin ^{a ~\- b): 
cos -J-c sin ^(^4 -{- B) = cos iO cos ^{a — I): 
cos ^c cos i{A -\- B) = sm iO cos ^{a + h). 
Proof. From the formulae for sums and differences of 

angles we have 

sin \{A ± B) = sin ^A cos ^B ± cos iA sin ^B; {a) 
cos ^{A ± B) = cos ^A cos ^B =F sin ^A sin ^B, (b) 



(13) 



98 SPHEBIGAL TBIGONOMBTRT. 

From (5) and the corresponding equations for the angles 
B and O we have 

. 1 . ./sin (s — h) sin is — c) 
^ sm sm c 



_ Vsin {s — c) sin (s —a)^ 

cos 1^^ 

cos ^B 



sin-J. 

sm c sm 0^ 

_ , /sin 5 sin (5 — 0^)^ 

~ ^ sin J sin c ' 

_ i/sin s sin {s — i) ^ 



sm c sm a 



• -1 ^ i/sin is — a) sin (5 — l) 

sm iC/ == y ^ — r— ^-^ — ^ ^; 

^ BYR a sm 6> ' 

-. /^ i /sin 5 sin is — c) 

cos iC = y : ^i — 7 — -' 

' sm ^ sm 

From these equations we find 

. \ . 1 73 sin (s — ^) .^ 

sm t^ cos ^B = ^ -^ cos tO: 

^ sm c ^ ^ 

sin (5 — a) , „ 
cosi^ sm \B = ^ cos -1(7: 

cos -1^ cos \B — - — sin \0\ 
^ sm c ' 

. sin (5 — c) . . „ 

sm tJ. sm t^ = ^ '- sm to. 

"^ sm c ^ 

By substitution in (a) and (J) we find 

sin i(^ -B)= cos ^c' ^i"(^-^)-«m(^-« ) 
^^ ^ sm (? 



^ 2 cos |-(25 — a — h) sin |-(6« — Z*) 



— cos iU o • 1 1 

2 sm ^c cos fc 

The value of 5 = -J-(a + & + c) gives 'ils — a ~ l = c. 
Thus we have, after multiplying by sin ^c, 

sin -Jc sin -i(^ — B) = cos i(7sin -^^ — Z>), 

which is the first of the equations (12). The other three are 
derived by a process so similar that the student can follow it 
through from the above model. 

The equations (12) are called Gauss's Equations, from 
the name of the mathematician who first brought them into 
common use. 



SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 99 

Example of Computation. 

Given a = 132° 46'.7, l = 59° 50M, C = 56° 28'.4. 

Required A, B and c, 

a, 132° 46'. 7 sin ^c sin i(^ - B), 9.7191 

^, 59 50.1 sin ic cos i(vl - ^), 9.6723 

a + J, 192° 36'. 8 tan i{A - B), 0.0468 

a -h, 72 56 .6 sin^c, 9.8475 

cos ic sin ^{A + B), 9. 8503 

K^ + Z*), 96° 18'.4 cosjc cos i(^ +i?), -8.7158 

i(a - Z*), 36 28 .3 

i(7, 28 14 .2 tan i(^ + B), - 1.1345 

cos |c, 9.8515 
tanic, 9.9960 
sini(rt-J), 9.7741 ^{A - B), 48° 5' 

cosiC; 9.9450 i{A-\-B), 94 12 

cos4(^-J), 9.9053 ic, 44 44 

sini(« + Z*), 9.9973 A, 142° 17' 

sin-i-(7, 9.6750 B, 46 7 

cosi-(6X + Z'), -9.0408 c, 89 28 

By this method sin ^c and cos -Jc are obtained by separate 
equations, and, if the work is correct, the two functions should 
correspond to the same angle. The quotient (diff. of loga- 
rithms) is tan ^c, from which we find ^c, and then see whether 
the sine and cosine of ^c agree with those computed. 

A still further check which should be applied is that of 

,. sin a &m h ., sin c , .„ ,, 

computmff-; — -., —. — 7^ and - — -^^ to see if they a^ree. 
■^ *sin^' sm^ sin C -^ ^ 

sin a, 9.8657 sin d, 9.9368 sin c, 0.0000 

sin^ 9.7866 sin^, 9.8578 sin (7, 9.9210 



0.0791 0.0790 0.0790 

EXERCISES. 

Compute the remaining parts of triangles of which parts 
are given as follows: 

1. Two sides are 28°20'.3 and 112° 1'.9; inc. angle 79°28'.6 

2. '' " '' 112 12.6 and 124 48.2; ^^ '' 18 17.0 

3. " '' '^ 19 59 .2 and 50 7 .6; '' '' 127 37 .8 



100 



SPHERICAL TRIGONOMETRY, 



(13) 



%^, Case IV. Given two angles and the side between 
tliem. Required the three remaining parts. 

This case may also be solved by Gauss's equations. To 
put them into a conyenient shape for this case the two mem- 
bers of each should be interchanged, and the order altered, 
thus: 

sin \C sin \{a -}- ^) = sin \c cos \(^A — B);'^ 

sin ^C cos i{a -{-!)) = cos -Jc cos i(A + -^); 

cos -1(7 sin i{a — i) = sin ^c sin i{A — B); 

cos iO cos i{a — h) = cos ^c sin ^{A + B). 

The form of computation varies so little from that of the 
preceding case that the student can readily construct it him- 
self. 

EXERCISES. 

Compute the remaining three parts of the following tri- 
angles: 

l.A= 32°68'.5, B= 65° 26'.7, 

B = 129 52 A, 

B = 67 58 .1, 

B = 55 11 .8, 



2. A = 160 .8, 
d. A = 151 39 .7, 
4:. A = 67 47 .4, 



c = 56° 21'.2. 
c = 52 22 .2. 
c = 100 31 .4. 
c = 161 43 .0. 



89. Solution of Cases III. ai^d IV. when alltheee 
OF the eemain^ing pakts aee not wanted. 

JVapier^s Analogies. If, in Case III., the side opposite the 
given angle is not required, but only the two remaining angles, 
we may eliminate that side from Gauss's equations (12). 
Dividing the first of these equations by the second, and the 
third by the fourth, we have 

sin i{a — b) 



tan i{A - B) 



sin ^{a 



tani(4 + i?) = °^^M^ 



cot ^C', 
cot -JC; 



(14) 



cos i{a -\- h) 
from which we readily compute A and B. 

Proceeding in the same way with equations (13), we have 

tan i(« + &) = ?5||i4-T-i tan ^c ; 



tan ^{a — h) 



co^i[A-\-B) 
sin i{A - B) 



tan ^c ; 



(15) 



sini(^ + ^) 
from which we may find a and l when A, B and c are given. 



SOLUTION OF SPHEBICAL TRIANGLES IN OENERAL. 101 

The four equations (14) and (15) are called Napier^ s Ana- 
logies, 

EXERCISES IN APPLYING NAPIER'S ANALOGIES. 

1. Given A = 135° 5'.5, B = 50° 30M, c = 69° 34:'.9. 
Eequired a and b. 

2. Given A = 65° 33', B = 95° 38', c = 100° 49'.5. 
Eequired a and i. 

3. Given a = 97° 27', b = 95° 38', C = 64° 23'. 
Eequired A and B. 

4. Given « = 82° 33', d = 84° 22', C = 115° 37'. 
Eequired A and ^. 

90. If only the side opposite the given angle in Case III., 
or the angle opposite the given side in Case IV., are required, 
we may use the last equations of sets (6) to (11), § 77. For 
instance, the last equation of (8) gives- cos c in terms of a, d 
and a 

If one angle is also required, we may employ either the 
first pair or the second pair of each set, according to the way 
in which the lettering is applied to the parts. 

In the form in which the equations are written in § 77 
the computation is longer than is necessary; we may shorten 
it by the following process: 

The parts a, d and C being given, we compute the auxi- 
liary quantities K and log k from the equations 

^ sin ^ = sin a cos C:) / x 

K cos Ji = cos fl^. 3 

By substituting these values in the second and fifth equations 

of (8) they become 

sine cos J. = ^cos^sin 5— /^sin^cos l = hsm (b—K); \ /,>. 

cos c = ^ cos j^cos b-\-k sin Ksinb = Jc cos (b—X). f 
Adding the first equation of the set unaltered, we have the 
three equations 

sin c sin ^ = sin C sin a, ^ 
sin c cos ^ = h sin {b — K), > {c) 

cos c = ^ cos {b — K),) 
to determine the two angles A and c. The quotient of the 
first two equations is 



102 8PHERIGAL TRIGONOMETRY. 

, sin C sin a 
tan A 



from whicli we determine A, Then we find sin c from either 

of the equations 

sin C sin 0^ h sin ih—K) 

sm c = : — -. — = ^^ — -. — -, 

sm A cos A 

and cos c from the last equation. These functions should, if 
the work is correct, correspond to the same yalue of c. The 
best course is to compute tan c = sin c : cos c, and thence c, 
and see whether the sine and cosine of c then found agree 
with the computed values. 

Example of Computatioi!^. 

To find the side c and the angle A for the example of § 87, 
in which were given 

a = 132° 46'.7; & = 59° 50M; C = 56° 28'.4. 



sin 0, 

sin a, 

cos G, 

cos a = h cos K, 

Jc sin K, 


9.9210 

9.8657 

9.7422 

- 9.8320 

9.6079 




1 

9 


sin {]) - K), 

log^, 

cos (Z» — K), 

sin c sin A, 
sin c cos A, 

tan A, 

A, 

sin c, 

cos Cy 

tan c, 


- 0.0000 
9.8982 
8.0669 


9.7867 
— 9 8982 


tan K, 


- 9.7759 

149° 10'. 
59 50. 




- 9.8885 
142° 16'.5 
0.0000 
7.9651 
2.0349 
89° 28'.3 


h-K, 


- 89° 19'. 



EXERCISES. 

1. Prove the following formulae for finding an angle and 
a side when two angles {A and B) and the side between them 
are given:, 

h sin IT = cos B; 
h cos II = sin B cos c; 
sin C sin b = sin B sin c; 
sin G Qosh — li cos {A — H)', 
cos G =^ h sin {A — H). 



SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 103 

2. What changes are required in the formulae {a), {I) and 

{0), 

First, when the parts sought are called B and c? 
Secondly, when the given parts are called l, c and A, and 
the parts sought are a and G ? 

3. Find the parts C and a in the first two exercises of § 88. 

91. Case V. Given two sides and an angle opposite one 
of them. 

Let the given parts be a, I and A. 
From the first equation (9), § 77, we find 

. „ sin A sin 5 
sm B — : , (16) 

from which B may be found. 

We may then find the parts c and G from Napier's Analo- 
gies. The equations (15) give 

+ar. 1. - COS i(^ + B) tan i{a + h) 

tan ,c - ^os i{A - B) ^ 

_ sin ^{ A + B) ta n i(a - l) 

~ &m^{A-'B) ' ^^^^ 

either of which may be used to find c. Jl both are used, the 
results will agree when the work is correct. 
For G we may use either of the equations 

sin A sin c sin B sin c 



sin G = 



sin a sin h 



but if G is near 90°, the result will be uncertain, because a 
small error in the sine will then cause a large error in the 
angle. (For example, by looking into the tables of trigono- 
metric functions it will be seen that if the log sine is 9.9998, 
the angle may be anywhere between 88° 4' and 88° 29'.) 
Hence in this case it is better after finding B to use one of 
the following equations derived from Napier's Analogy (14) : 

tan 1(7 = sin ^{a - h) cot ^(A - B ) 
^ sin ^{a + h) 

_ cos i{a - h) cot ^{A + B ) 

~ cos i{a -{-])) ~ ' ^^^^ 



104 SPHERICAL TBIGONOMETBT. 

92. Case VI. Given two angles and a side opposite one 
of tliem. 

This case may be solved by almost the same equations as 
the last one. If we call the given parts Aj B and a, we have 

. , sin a sin B 

sm = : -. — , 

sm A 

from which we find d. We then find c and C from (17) and 

(18). 

EXERCISES. 

Find remaining parts when — 

1. Given i = 85° 52'; c = 75° 52'; B = 84° 14'. 

2. Given b = 58 19 ; c = 42 2 ; (7 = 19 57 . 

3. Given a = 155 22 ; ^ = 158 30 ; B = 58 14 . 

4. Given b= 96 62 ; B = 97 7 ; (7 = 75 19 . 

MISCELLANEOUS EXERCISES IN SPHERICAL. 
TRIGONOMETRY. 

1. From the vertex A of an isosceles spherical triangle in 
which AB = AC = Sr ^T and BG = 132° 53', an arc of a 
great circle is drawn to a point D on the base BG, such that 
GD = 59° 15'. Find the length of this arc. 

2. What are the face- and edge-angles of a triangular 
pyramid each of whose faces is an equilateral triangle? 

3. Four identically equal triangles the sides of each of 
which are 3, 4 and 5, are joined together so as to form a tri- 
angular pyramid. Compute the face- and edge-angles of this 
pyramid. 

4. The faces of a pyramid whose base is a level square are 
each inclined 30° to the vertical. Find the inclination of the 
edges and the angles of the triangles which form the faces. 

5. If four points, each equally distant from the other 
three, be taken on the surface of a sphere, how far apart will 
they be? Apply to the case in which the sphere is the earth, 
one point the north pole, and the other three points equaljy 
distant from each other and from the pole. In what latitude 
must the three points be placed? 

6. The Washington Observatory is in latitude 38° 53'.6 N., 
the Greenwich Observatory in latitude 51° 28'. 6 N., and their 



SOLUTION OF SPHERICAL TRIANGLES IN GENERAL. 105 

difference of longitude is 77° 3'. What is their distance in 
miles, 69|- of which make a degree, assuming the earth to be 
a sphere; and what angle does the great circle from Washing- 
ton to Greenwich make with the meridian at Washington? 

Note. In this question we consider the spherical triangle of wKich 
one vertex, P, is the north pole, and the others, M and N, are the two 
points of the earth in the problem. We have for the parts of this tri- 
angle 

PM= 90° — north latitude of Washington; 
PN = 90° — north latitude of Greenwich; 
Angle P = difference of longitude. 

But if one or both the points if and iV were in the Southern Hemisphere, 
we should have 

90° + south latitude 
for the corresponding sides of the triangle. 

In either case we use the north-polar distances of the places as sides 
of the triangle. 

7. What is the distance from the Observatory of Greenwich 
to that of Kio de Janeiro, whose position is: 

Latitude - 22° 54'; 
Longitude 43° 10'. 

Note. In the Southern Hemisphere latitudes are considered as alge- 
braically negative. 

8. If a ship should sail from 'New York in a direction 
east 12° north, and continue on a great circle, at what dis- 
tance would she be found sailing due east? The latitude of 
New York is 40° 43'. 

9. If a ship should, sail southeast from Eio de Janeiro, 
and continue her course on a great circle, what would be the 
highest latitude she would reach? And where would she 
reach it? 



Land Sueyeting. 



[107] 



LAND SUEVEYING. 



lo Definitions, Surveying is the art of measuring, 
laying out and dividing land. 

It is diyided into three branches — Plane, Geodetic and 
Topographical. 

In Plane Surveying, the portion surveyed is regarded 
as a plane surface. 

In Geodetic Surveying, the curvature of the earth is 
taken into account. 

In Topographical Surveying, the irregularities of 
the surface, the course of streams, the position of ponds, 
lakes, mountains and other objects, are considered and de- 
lineated on paper. 

MeASUEEMEKT of LlisTES. 

2. The instruments employed in plane surveying for meas- 
uring distances are Guntefs Chain, so called from the name 
of the inventor; the Engineers Chain, and the Tape-measure. 

Gunter's Chain is 4 rods or QQ feet in length, and is 
divided into 100 links, each of which is therefore equal to 
7.92 inches. 

The Engineer's Chain, used chiefly in surveying rail- 
roads and canals, is 100 feet in length, and is divided into 100 
links. 

The Tape-measure is divided into feet and inches, and 
is used only for measuring short distances, such as the boun- 
daries of town-lots, garden-plots, etc. 

For measuring distances with the chain eleven iron pins, 
about 12 inches long and -J inch thick, are used for marking 
the terminations of each length of the chain. The person 
who goes forward, called the leader, takes one end of the chain 
and ten pins, and the other person, called the follower, takes 
the other end of the chain and the other pin which he 
pushes into the ground at the beginning of the line to be 
measured. When the leader has advanced the length of the 



110 LAND SUBYETING. 

chain, the follower directs him to the right or left until 
the chain coincides in direction with the line; the chain is 
then stretched tightly and held in a horizontal position. 

This being done, the leader sticks one pin into the ground 
and the follower takes up the pin at the beginning of the line. 
The chainmen then move forward and measure a second 
chain in the same manner, and so on until all the pins are 
exhausted, when a record that ten chains have been measured, 
is made by the follower. Ten pins are then returned to the 
leader, the other one remaining in the ground to mark the 
beginning of the next chain, when the measuring proceeds, 
as before, until the whole line is measured. In measuring up 
or down an inclined surface, such as the side of a hill, the 
chain must be maintained in a horizontal position; and when 
the incline is considerable, a fraction of a chain may be used. 

3. Obstacles to Cliai7iing. In measuring or extending a 
line, obstacles such as houses, trees; rocks, ponds or rivers 
are sometimes met with, and may be passed in various wa3^s. 
Thus, suppose we want to pass the obstacle in the line PQ. 
At any convenient points A and B lay off the equal perpen- 

P A E r>\ G Ho 



<»> 



CD E P . 

diculars AG, BD, which for small objects, such as a tree or 
a house, need not exceed 20 or 30 links, and extend the line 
CD to two other convenient points E, F, at which lay off the 
perpendiculars EG and FH, each equal to AG ov BD-, then 
6^^ will evidently be in the same straight line as AB, and the 
distance BG will be equal to DE, which can be measured. 

When a large obstacle is met with, such r.s a pond or river, 
it may be passed by triangulation, as follows: 

Suppose we want to extend the line PB across the pond 
LK. At any convenient point in the line, as A, measure off 
AB equal to 40 links; fasten one end of the chain at B by 
means of one of the chaining-pins, stretch the chain at right 
angles to AB as nearly as can be estimated, and locate the 
point G so that BG may be equal to 30 links and ^(7 equal to 
60 links. The angle ABG m\\ be a right angle, for, by the 



MEASUREMENT OF LINES AND ANGLES 



111 



construction, AC = AB' + BC\ Extend the line AG to 
some point D such that when DH is drawn perpendicular to 

J. 

/ 

Q 



h 




AD it may clear the obstacle; then measure off Db equal to 40 
links, and locate the point c so that Dc may be equal to 30 
links and ic 50 links, and erect a flag-staff at some point H 
in Dc produced and beyond the line PQ. From the similar 
triangles ABC, ADE y^Q have 

AB: BC::AD:DE, 
or 40 : 30 ::AD : DE-, 

whence DE = ^AD, 

Measure off now the line DcE, equal to f of AD, and the 
point E, in the line PB produced, is established. 

Again, AB\AC::AD:AE, 

or 40 : 50 :\AD : AE-, 

whence AE = ^AD. 

If the point ^ is not visible from any of the points P, A 
or B, we must establish in the same manner another point F; 
then EF will be in the same straight line with AB. 

Measurement of Angles. 

4. The instruments employed in surveying for measuring 
angles are the Magnetic Compass and the Theodolite or Tran- 
sit Tlieodolite. 

The Surveyor's Compass is illustrated on page 112. 

The needle turns freely on a pivot at the centre, and when 
at rest coincides in direction with the magnetic meridian. 



112 



LAND SUBVETING. 




THE SURVEYOR'S COMPASS 113 

The com2)ass-circle, within which the needle turns, is 
graduated on its upper surface to half-degrees, and numbered 
from. 0° to 90° on each side of the north and south line. 

Tlie sight-standards f a, h, are firmly fastened at right 
angles to the plate by screws shown in the drawing, and are 
perforated through nearly their whole length by slits which 
are terminated at intervals by circular apertures through 
which the objects toward which the sights are directed, may 
be more easily found. Two spirit-levels, c, d, placed at right 
angles to each other, are attached to the compass-plate. The 
right and left edges of the north sight-standard are graduated; 
the former is used for measuring angles of eleyation, and the 
latter for angles of depression. 

Eye- pieces are placed on the right and left sides of the 
south sight-standard, each of which is on a level, when the 
compass is level, with the zero of the graduated or tangent 
scale on the other sight-standard. These eye-pieces are 
centres of the arcs to which the north sight-standard is tan- 
gent at the zero-point. 

The instrument when in use is supported by a tripod or 
sometimes by a single staff about 5 feet long, having at the 
upper end a ball-and-socket joint, and terminated at the 
other by a sharp steel point, so as to be firmly placed in the 
ground. The joint is furnished with a clamp-screw by which 
the instrument may be securely held in any position. 

When the instrument is not in use, a concealed spring, 
called the needle-lifter (not shown in the drawing), lifts the 
needle from off the pivot. 

Near the south sight-standard the compass-plate is gradu- 
ated for a few degrees to 30' spaces, and on the concave side 
of the graduated arc and attached to the compass-box there is 
a scale called a Vernier, from the name of the inventor, for 
measuring fractional parts of the divisions. A clamp-scrcAV 
holds the compass-box firmly against the compass-plate, and 
the tangent-screw {t) serves to give the former a slow motion 
on the latter. 

Principle of the Vernier, 

5. Let ^^ be an arc graduated to half -degrees or 30', and 
VV the vernier^, so graduated that 30 spaces of the vernier 



114 



LAND SURVEYING. 



are equal to 29 spaces on the limb; that is, one space on the 
vernier is equal to 29', and therefore the difference between 
one space of the limb and one of the vernier is 1'; hence jt 
follows that when the first division-line of the vernier coin- 




cides with a line on the limb, the zero-point of the vernier is 
1' from the nearest division-line on the limb. When the 
second line of the vernier coincides with a line on the limb, 
the zero of the vernier is 2' from the nearest line on the limb, 
and so on. In order, then, to read an angle for any position 
of the vernier, we pass along the vernier until we find a line 
coinciding with a line of the limb; the number of this line, 
counted from the zero-point, indicates the number of minutes 
which must be added to the degrees and half-degrees marked 
on the limb. Thus, the reading of the vernier in the drawing 
is 52° 9'. 

The vernier is here supposed to be moved in the direction 
in which the graduations are numbered; but if it be moved in 
the opposite direction, the number of minutes obtained as 
above must evidently be subtracted from 50' to obtain the 
correct reading. 

In the compass, however, the vernier is made double, as 
shown in the diagram on page 115. It has only 15 spaces on 
each side of the zero-line, and is numbered in two series; the 
lower in the diagram and the one nearer the observer, who is 
supposed to stand at the south end of the instrument, is 
numbered 5, 10, 15 each way from the zero-point (0); the 
upper series has 30 over the zero-point, and 25, 20, 15 each 
way above the 5, 10 and 15, respectively, of the lower series. 
When the vernier is moved to the right, read the lower series 



THE VERNIER. 



115 



to the left or the upper series to the right of the zero-point, 
and vice versa. 




The chief use of the vernier in the compass is to set off 
the variation of the needle. When the variation is known at 
any place, the tangent-screw and vernier enable us to move 
the compass-circle through an arc equal to the variation. If 
the variation is west, the vernier is moved to the right, and 
vice versa. 

Adjustments of the Compass. 

6. The adjustments of the compass are three in number: 
1st. To adjust the level; 
2d. To adjust the sights; 
8d. To adjust the needle. 

The last two are made by the maker and seldom require 
any attention from the observer. 

1st. To adjust the Level. 

Bring the bubbles to the centre of the tube by gently 
pressing the plate; turn the instrument half round; then if 
the bubbles run to one end of the tube, that end is the higher. 
Loosen the screw under the lower end and tighten the one 
at the higher end until the bubble is brought half-way back. 
Level again and repeat the operation until the bubbles remain 
in the centre during an entire revolution of the instrument. 

2d. To adjust the Sights. 

Perform the first adjustment, and then observe through 
the slits a fine thread made perpendicular by a weight which 
must 'be immersed in a vessel of water in order to prevent its 
vibration. If both sights do not exactly coincide in direction 



116 



LAND SUBVETING. 



with the thread, correct them by filing off a little from the 
under surface of the higher side. 

3d. To adjust the Needle. 

Perform the first adjustment, and by means of the tangent- 
screw and vernier make one end of the needle coincide with 
any division of the circle, and observe if the other end is 180° 
from the first; if not, move the centre pin or pivot, by means 
of the screws holding it to the compass-box, until the ends of 
the needle mark opposite degrees. Now turn the compass 
half-way round, and again observe whether the needle marks 
opposite degrees; if not, correct half the error by bending the 
needle, and the other half by moving the centre pin, and 
repeat the operation as often as necessary. 

Uses of the Compass, 

•y. Definitions. The leaving of a line is the angle it 
maTces with the meridian passing through one extremity ^ and 
is rechoned from the north or south point of the horizon to- 
ward the east or west. 

The magnetic hearing of a 
the magnetic meridian. 

Thus, if N8 be a meridian 
either true or magnetic, and 
the angle BAN= 40°, then the 
bearing of the line AB from 
A is 40° to the east of north, 
or, as usually read, *^ north 40' 
east," and written N. 40° E. 
The reverse bearing of a line 
is the bearing taken from the 
other end; thus the bearing 
of the line AB from B is 
S. 40° W. 

To taJce the Bearing of a Line. 

Place the compass on the line, turn the north end in the 
direction of the course, and, standing at the south end, direct 
the sights to some well-defined object, as a flag-staff, in the 
course. The north end of the needle will, when it comes to 
rest, indicate the bearing to the nearest half-degree; if the 



is the angle it mahes with 



W 




E 



VABIATION OF THE NEEDLE. 117 

bearing to minutes is required, the vernier must be used. 
On the face of the compass the letters E and W are reversed 
from their true position. This is done to facilitate the taking 
of bearings; for if the sights are turned towards the east, the 
north end of the needle will be directed toward E, which 
therefore indicates that the bearing is east of the magnetic 
meridian, and vice versa. 

From a given Point to run a Line having a given Bearing, 

Place the compass directly over the given point, and 

turn the compass until the needle indicates the given beariDg, 

using the vernier if necessary; the line of sights observed 

from the south end will be the required line. 

Variation of the Needle. 

8. Def. Tlie variation of the needle is the a7igle which the 
magnetic meridian mahes toith the true meridian, and is east 
or west according as the north end of the needle is east or west 
of the true meridian. 

The variation is different in different localities; nor is it 
constant at any one place. The line of no variation is the 
line passing through those points on the earth's surface where 
the needle points due north and south. At all places east of 
this line the variation will be west, and at all places west of 
this line the variation will be east. West variation is regarded 
as + (positive), and east variation as — (negative). 

The needle is subject to irregular changes coincident with 
thunder-storms, auroral displays, sun-spots, etc. 

Diur7ial Changes. In the northern hemisphere the north 
end of the needle moves from 8' to 12' west from about 8 
A.M. to 2 P.M., and then gradually returns to the original 
position. The diurnal change varies with the season, being 
nearly twice as great in summer as in winter. 

Secular Changes. This is a change in the same direction, 
extending, with considerable regularity, through a period of 
about 150 years, and then in the opposite direction for the 
same length of time. 

In this country the line of no variation, or the magnetic 
meridian, extends at present (1884) through Michigan, Ohio, 
Virginia and the Oarolinas, and is gradually moving west- 



118 LAND SUBVBYma. 

ward, so that afc places east of this line the variation is increas- 
ing, and decreasing afc places on the west side. 

The following table gives the variation for several places 
on this continent for the years 1880 and 1885 : 

Annual 

Variation, Change. 

Lat. Long. 1880. 1885. 1880. 

Boston, Mass 42".4 71°.0 11°.56 11°. 78 2'.9 

West Point, K Y. . . . 41°.4 74°.0 7°. 80 8°.00 2'.4 

Cleveland, Ohio 41°.5 81°.7 1°.31 1°.51 2'.5 

Detroit, Mich 42°.3 83°.0 0M3 0°.37 3'.0 

Charleston, S. C 32°.7 7r.9 - 0°.41 - 0M7 3'.0 

Savannah, Ga 32°. 81°.0 - 1°.89 - 1°.60 3'. 5 

New Orleans, La 29°.9 90°.0 - 6°.49 - 6°.18 3'. 5 

San Francisco, Cal.... 37°.8 122°.4 -16°.52 -16°.56 -0'.5 

9. The Surveyor's Transit, or the Transit Theodolite, 
combines the advantages of both the simple transit instrument 
and the theodolite, and is shown in the annexed drawing. It 
consists essentially of two concentric circular plates of brass 
having a horizontal motion about a vertical axis. The lower 
plate is graduated to half-degrees, and is called the graduated 
limb of the instrument; the upper is called the vernier-plate, 
and supports the compass-box, the supports for the telescope and 
its attachments. The axis of the graduated limb is perforated 
longitudinally to receive the axis of the vernier- plate, so that 
the latter turns freely on the former. The graduations are 
numbered from 0° to 360°, and are read by a vernier to single 
minutes. 

The parallel plates A, B are held together by a ball which 
rests in a socket, and on the upper surface of the lower plate 
four screws turn in sockets firmly attached to the plate and 
have their heads press against the under surface of the upper 
plate. By means of these four screws the instrument is 
levelled. On the upper side of the upper plate are a clamp- 
and tangent-screw. By loosening the clamp the graduated 
limb may be turned on its axis to any desired position, and 
when clamped the tangent-screw serves to give it a slow mo- 
tion. To the vernier-plate, or its attachments, two levels a, T), 
are attached at right angles to each other, for the purpose of 
making the graduated limb horizontal, and a clamp- and 



THE TRANSIT THEODOLITE. 



119 




The Transit Theodolite. 



120 LAND STJUVETING, 

tangent-screw (shown at D) serve to clamp the vernier-plate 
and give it a slow motion. 

The telescope T, is supported by an axis which rests on 
supports firmly fixed to the vernier-plate and high enough to 
permit the telescope to make an entire revolution on its axis. 
The vertical circle cd, graduated to half -degrees and having 
its plane parallel to the optical axis of the telescope, is attached 
firmly to one end of the axis on which the telescope revolves, 
and is read by the vernier v, to single minutes. The axis of 
the telescope is also supplied with clamp- and tangent-screws 
for fixing the telescope in any position and giving it a slow 
motion. 

Under and parallel to the optical axis of the telescope a 
spirit-level is placed to show when the telescope is brought 
into a horizontal position. For the purpose of enabling us 
to direct the telescope upon an object with precision, two 
spider-lines or very fine platinum wires are fixed, one vertical 
and the other horizontal, in the focus of the telescope. These 
wires are fastened across a diaphragm smaller in diameter 
than the tube of the telescope, and held in it by four screws 
passing through the tube. By loosening one of these screws 
shown at s, and tightening the opposite one, the diaphragm 
may be moved laterally or horizontally. 

Adjustments. 

10, 1st. To make the Vernier-plate Horizontal. 

Place the levels on the vernier-plate parallel to the lines 
which pass through opposite pairs of levelling-screws, then 
turn the screws in opposite directions until the bubbles are 
brought to the centre of the tubes. Now turn the vernier- 
plate through 180°, and if the bubbles remain in the centre 
the adjustment is made; but if not, correct half the error 
by the screws which hold the levels to the vernier-plate, 
and the other half by the levelling-screws. Owing to the 
difficulty of estimating exactly one half the error, this process 
will have to be repeated until the bubbles remain in the centre 
of the tubes, while the vernier-plate makes a complete revolu- 
tion on its axis. 

2d. To adjust the Line of Collimation, 



TEE TRANSIT THEODOLITE. 121 

This adjustment consists in bringing the intersection of 
the wires into the optical axis of the telescope^ and is per- 
formed as follows : Make the first adjustment, and then 
draw out the eye-piece until the wires are clearly seen. When 
this is the case, a slight movement of the observer's eye pro- 
duces no apparent displacement of the wires as they are seen 
projected on the object. The object-glass is then to be drawn 
out by the rack and pinion until the object is distinctly seen. 
The wires are then in the common focus of the object-glass 
and eye-piece. This operation should be repeated whenever 
the distance of the object is changed, especially if it is near. 
Place the intersection of the wires on some distant and well- 
defined object; clamp the instrument, and having revolved 
the telescope half round on its axis, find or place some object 
in the opposite direction at about the same distance from the 
instrument as the first object. In turning the telescope great 
care should be taken not to disturb either the graduated limb 
or the vernier-plate. 

Having found an object, place the intersection of the wires ^ 
on it, unclamp the instrument, turn it half round, and direct 
the telescope to the first object, and having placed the inter- 
section of the wires on it, again clamp, revolve the telescope 
as before, and observe if the intersection of the wires now 
bisects the second object; if it does, the wires are adjusted; 
but if not, half the error must be corrected by the screws 
which move the diaphragm, and the operation repeated until 
perfect coincidence takes place in both positions of the tele- 
scope. 

3d. To make the Axis of the Telescope Perpendicular to the 
Vertical Axis of the Instrument. 

This adjustment requires that the intersection of the wires 
may trace a vertical circle as the telescope is revolved on its 
axis. Having previously adjusted the line of coUimation, 
level the instrument near some lofty object, such as a tower 
or spire, and fix the intersection of the wires on some well- 
defined point on the top of the object; clamp both the gradu- 
ated limb and vernier-plate and bring the telescope down till 
the wires bisect some good point found or made at the base; 
turn the instrument half round and fix the wires on the lower 



122 LAND SURVETINa. 

pointy clamp again and raise the telescope to tlie liigher point: 
then if the wires bisect it, no further adjustment is necessary. 
If, however, the wires are thrown to one side, the end of the 
axis opposite to that side is higher than the other and must 
be lowered by the screw which eleyates or depresses the end 
of the axis. The operation is then to be repeated until per- 
fect coincidence takes place. 

Or thus, haying performed the first two adjustments, ob- 
serve if the vertical wire coincides with the plumb-line sus- 
pended at some distance in front of the instrument as the 
telescope is raised or lowered. If it does, the intersection of 
the wires will describe a vertical circle and the adjustment is 
already made; but if not, make the correction as above indi- 
cated. 

4th. To make the Axis of the Level attached to the Telescope 
Parallel to the Line of Colli^nation. 

Having made the first four adjustments, bring the telescope 
to a horizontal position as nearly as can be estimated by the 
eye. At the distance of two or three hundred feet drive a 
stake into the ground and observe the height indicated by a 
staff set perpendicularly upon the top of the stake. Clamp 
the telescope and turn the instrument through 180°, and at 
the same distance from the instrument drive another stake 
into the ground. Place a staff perpendicularly on this stake 
and drive it down until the same height is indicated as in the 
first observation. The top of the stakes will now be in the 
same horizontal line whether the telescope is horizontal or not. 
Place the instrument at a point in the line joining both stakes, 
and on the same side of both of them, and at about one or two 
hundred feet from the nearer one; level the instrument, and 
move the telescope by the tangent-screw until the horizontal 
wire indicates the same height on a staff placed on both 
stakes. The telescope will then be horizontal, and without 
disturbing it bring the bubble to the centre by the screws at 
the ends of the level, which will complete the adjustment. 

The adjustments are always made by the maker of the 
instrument; but notwithstanding this, every instrument re- 
quires to be readjusted frequently. The needle is adjusted in 
the same manner as in the compass. 



TEE TRANSIT TEEODOLITE. 123 

Uses of the Trmisit TJieodolite. 

11. The transit theodolite is used for measuring horizon- 
tal and yertical angles, and with much greater precision than 
can be done with the compass. 

To measure a Horizontal Angle. 

Place the instrument directly over the vertex of the angle 
to be measured, level the instrument, which is of course sup- 
posed to be accurately adjusted, set the zero of the vernier at 
zero on the limb, and clamp the vernier-plate; unclamp the 
limb and turn the telescope in the direction of one of the sides 
of the angle, and by means of the tangent-screw make the line 
of collimation coincide exactly with the side of the angle; now 
clamp the limb and unclamp the vernier-plate, and turn the 
telescope in the direction of the other side, using the tangent- 
screw of the vernier-plate to make exact coincidence. The 
reading indicated by the vernier is the angle required. Some- 
times the vernier-plate has two verniers, 180° apart; in that case 
the mean of the readings of the verniers should be taken, as 
in this way any error of eccentricity in the limb is corrected. 

To measure a Vertical Angle. 

Level the instrument, and bring the bubble of the level 
attached to the telescope to the centre of the tube by the 
tangent-screw; the line of collimation will then be horizontal, 
and the zero of the vertical circle and its vernier must be made 
to coincide. Then direct the telescope to the object whose angu- 
lar elevation is required, and clamp the vertical circle. The 
reading of the vernier will indicate the angle of elevation. 

The vernier of the vertical limb is double, and is read in 
the same way as the vernier on the compass, already described. 

To Establish a Teue Meridiak. 

12. 1st. By olservation of Polaris wlien on the meridian. 
The north star {a Urs83 Minoris), or Polaris, is at the pres- 
ent time (February, 1884) 1° 18' 14".5 from the pole around 
which it apparently revolves in a sidereal day, or 23^ 56"^ 4'.09 
of mean solar time; and in this period it will in general cross 
the meridian twice, once above the pole, called the upper cul- 
mifiation, and 11^ 58^ 2^04 later below the pole, called the 
lower culmination. It arrives at its greatest distance from 



124 



LAND SURVEYING. 



the meridian, or its greatest eastern or western elongation, 
about 5^ 59"^ before or after culmination. The following table 
gives the time of upper culmination to the nearest minute for 
the year 1885: 



1st. 

January &" 35"" P.M. 

February 4 33 " 

March 3 38 " 

April 12 32 " 

May 10 34 a.m. 

June 8 33 " 

July 6 35 " 

August 4 34 '• 

September 2 32 " 

October 12 35 " 

November 10 33 p.m. 

December 8 35 " 



nth. 




21st. 


5^56°^ 


P.M. 


51^ 


16" P.M. 


3 53 


" 


3 


14 '* 


1 59 


<( 


1 


20 *' 


1 53 


A.M. 


11 


14 A.M. 


9 55 


" 


9 


16 " 


7 54 


<< 


7 


15 '.' 


5 56 


<t 


5 


17 " 


3 55 


" 


3 


16 " 


1 53 


«< 


1 


14 " 


1 55 


P.M. 


11 


16 P.M 


9 53 


<( 


9 


14 " 


7 55 


<< 


7 


16 '' 



>7 



■^ Tolaris 



PoU 



Alioth ^_ 



The time of upper culmination may be found approxi- 
mately by means of the star Alioth {s Ursas Majoris) of the 
constellation of the Great Bear, which is nearest the four 
stars forming the quadrilateral. 

Suspend from a height of 20 feet or 
more a plumb-line. The bob should 
be in a vessel of water to prevent 
its vibration. Then, about twenty 
minutes after Polaris and Alioth ap- 
pear to coincide with the plumb- 
line, Polaris will be on the meridian. 
Direct the telescope of the transit 
theodolite to the star a few minutes 
before the time of culmination, fol- 
low the star by the tangent-screw of 
the vernier-plate, and at the instant 
of culmination clamp the instrument. % 
The line of collimation of the tele- 
scope will then be in the plane of the 
meridian which may be traced out 
on the ground by bringing the tele- 
scope down to a horizontal position. 

2d. By observation of Polaris or any other circumpolar 
star when at its greatest elongation. 



^fi 





AZIMUTH OF A STAB. 125 

Def. The azirnuth of a star is the angle which the vertical 
plane through the star makes tuith the plane of the meridian. 

Let NZO be the merid- Z 

ian, Z the zenith, and NO 
the horizon of an observer; 
F the north pole of the heav- 
ens, and S the position of 
the star at its greatest east- 
ern elongation. Then PiV^oL^^ _ ^^^ 
is the latitude of the ob- 
server, P>Sthe star's polar distance, and PZS its azimnth. Put 
PN = cp, FS = p, and we have from the spherical triangle 
FZ8, by spherical trigonometry, 

sin ZF : sin SF :: sin FSZ : sin FZS, 
or cos ^ : sin ^ : : sin >S^ : sin Z; 

whence sin Z = — sin S. (1) 

cos cp ^ ^ 

Now it is evident that the azimuth (Z) is greatest when 

the star is at its greatest elongation; and since is, for 

^ to ? COS ^ ' 

the same place and time, a constant quantity, sin Z will be a 
maximum when sin S has its greatest value — that is, when 
S = 90°; therefore at the time of greatest elongation the 
triangle ZF8 is right-angled at S, sin S = 1, and the above 
formula then becomes 

. ^7 sin p ,_. 

sm Z = ~, (2) 

cos q) ^ ^ 

The following table gives the north-polar distance of 
Polaris for the 1st of January of each year: 

Date, N. P. D. 

1884 1°18'14".4 

1885 l°ir 58".0 

1886 1°17'41".4 



Date. 


N. P. D. 


1887 


........ i°ir 24".i 


1888 


i°ir 6".2 


1889 


ri6'48M 



To find the Time of greatest Elongation. 

At the time of upper culmination the right ascension of 
the star (given in the American Ephemeris) is the sidereal 
time at the place. This sidereal interval converted into mean 
solar time will give the local mean time of the upper culmina- 



126 



LAND 8UBVETING. 



tion; then add 6^' 59°^ for the western or subtract 5^ 59°^ for 
the time of the eastern elongation. 

The times of greatest elongation may also be found with 
sufficient accuracy from the following tables, which will serve 
for the remainder of this century, the change of time being 
on an average only -f 19 seconds per year. 



Western Elongation. 



1st. 

January 12^ 29™ a.m. 

February 10 23 p.m. 

March 8 32 " 

October 6 32 a.m. 

November 4 31 " 

December .,...,... 2 32 " 



nth. 

11'' 45m pjyi_ 

9 43 " 
53 " 
53 
51 
53 



A.M. 



21st. 

9 4 " 

7 14 " 

5 14 

3 12 

1 14 



M. 



Eastern Elongation. 



1st. 
6^ 41"* A.M. 



April. 

May , 4 43 

June o.. 2 42 

July 12 44 

August 10 39 

September 8 37 



P.M. 



6^ 
4 
2 
12 
10 
7 



nth. 

2"" A.M. 

4 
2 
1 


58 



21st. 

n^ 23'" A.M. 

3 25 " 

1 23 " 

11 22 P.M. 

9 20 " 

7 18 " 



A few minutes before the time of elongation set up the 
instrument in a convenient place and level it carefully. 

About two or three feet north of the instrument fix to a 
vertical staff or post, a piece of board about one foot square, 
covered with white paper and perforated through the centre 
by a hole two or three inches in diameter. On the lower edge 
of this board fix some contrivance for holding a candle or 
small lamp, the light of which reflected from the paper will 
render the spider-lines visible. Direct the telescope to the 
star, and place the board high enough to allow the star to be 
seen through the orifice. As the star approaches its greatest 
elongation, follow it by the tangent-screw on the vernier- 
plate, keeping the vertical wire in range with the star. When 
the star arrives at its greatest elongation it will appear to 
coincide with the vertical v/ire for some seconds and then 
depart from it in the opposite direction. Clamp the instru- 
ment and bring down the telescope to a horizontal position. 



I)ET,EBMINATION OF AREAS. 



127 



Tlie azimutli of the star may now be turned ofi to the east or 
west according to the elongation observed, and the line of 
collimation of the telescope will then be in the plane of the 
meridian, which may be marked on the ground by driving a 
stake directly under the instrument and another 15 or 20 rods 
to the north of the instrument so as to range with the vertical 
wire of the telescope. The exact point on the stake may be 
marked by driving in a small nail or tack, using a lamp, of 
course, to render the stake visible through the telescope. Tlie 
exact point in the other stake should also be marked by driv- 
ing in a tack directly under the centre of the instrument as 
indicated by the plumb-line. 

Determination" of Areas. 

13. The area of a tract of land is the horizontal surface 
contained within its boundaries. The unit of surface is the 

Acre = 10 square chains = 100,000 square links = 4 roods 
= 160 square rods or poles. 

In making a survey of a tract of land, we begin at one 
corner and go entirely around it, keeping the field on the 
right. The lengths of the bounding lines and their bearings 
are entered in a book prepared for the purpose and constitute 
what are called field-notes. The left-hand page is ruled in 
three vertical columns, in the first of which the number of 
the station is entered, in the second the bearing of the side, 
and in the third the length of the side. On the right-hand 
page a rough outline of the survey is drawn with the bearings 
and distances indicated, as below. 

Form for Enterlng Field-Notes. 



Station. 


Bearings, 


Distances, 




OUTJ.INE. 








Ch. 


N- 


2! 




1 


N. 36° r E. 


29.17 




|/N>\ 


, ! 


2 
3 


S. 67° 15' E. 
S. 45°58'W. 


47.20 
53.77 


1 


S/ 




4 


N. 34° 32' W. 


38.92 


S 


V 





128 



LAND 8UBVETING. 



Angle letween Two Courses, 

14. The angle between any two courses, whether adjacent 
or not, is frequently required in plotting the survey and for 
checking the bearings. 

Keyerse one of the bearings if necessary, so that both 
courses may extend from the same point; then the angle be- 
tween any two contiguous sides is found as follows: 

1. If both courses are north or 
south, and both east or west, the 
included angle is equal to the differ- 
ence of the bearings. Thus the 
bearings oi AB and AC are respec- 
tively N. 48° W. and N. 10^ W.; 
hence the angle CAB = 38°. 

2. If one course is north and 
the other south, but both east or 
west, the included angle is equal to 
the supplement of the sum of bear- 
ings. Thus if AD bears 1^. 25° E. 
and^^' S. 48° E., then the angle DAB' 
= 107°. 

3. If both are north or south, but one east and the other 
west, the included angle is the sum of the courses. Thus the 
angle CAD = 35°. 

4. If one is north and the other south, one east and the 
other west, the included angle is the supplement of their dif- 
ference. Thus, AB is N. 48° W. and A C is S. 10° E.; then 

BAC = 180° - B'AC = 180° - (WAB -SAC'). 

Ex, Find the angles of the survey whose field-notes are 
given above, and show that their sum must be equal to four 
right angles. 

Latitude and Departure. 

15. Def, The LATiTTJB'Ei of a course in surveying is the dis- 
tance that one end is farther north or south than the other end. 

TJie departure of a course is the distance that one end is 
farther east or west than the other end. 




180' 



S' \C 

- (48° + 25°) 



LATITUDE AND DEPABTUBE. 



129 



Thus, \i AC be any course, then AD is its latitude and 
CD its departure. The distance, latitude and N 
departure constitute a right-angled triangle. 
If we denote the bearing by B, the course by c, ^ 
the latitude by I, and the departure by d, we 
shall have 



d = c ^m B 



:\ 



I-- 



(3) 




If a course runs due east or west, its latitude 
is 0; if due north or south, its departure is 0. -A. 
North latitude is considered + (plus); south 
latitude, — (minus); east departure, -f (pl^^s); 
and west departure, — (minus). 

When a survey has been correctly executed, and the latitudes 
and departures accurately computed, the sum of the north- 
ings should be equal to the sum of the southings, and the sum 
of the eastings to the sum of the westings. In practice, how- 
ever, this seldom happens, because the chaining is never per- 
fectly correct, and the bearings are taken only to minutes of 
arc. Small errors are therefore unavoidable; and when an error 
is as likely to occur in one course as in another and does not 
much exceed one link to three or four chains, the errors in 
latitude and departure are distributed among the courses 
according to their length by the following proportion: 

As the perimeter of the field : the length of one side :: error 
in latitude or departure : correction in latitude or departure 
for that side. 

If one side is more difficult to measure than any of the 
others, it may be fairly assumed that the error lies chiefly in 
that side, and the correction should be made accordingly; or 
if, in taking a bearing, the object could not be clearly seen, 
the error probably occurred there, and the correction applied 
to the latitude and departure of that course. When the 
corrections have been applied so that the sum of the north- 
ings is equal to the sum of the southings, and the sum of the 
eastings to the sum of the westings, the work is said to be bal- 
anced. 

Thus from the preceding field-notes we have the following 
table: 



130 



LAND SURVEYING. 





Distance. 


Latitude. 


Departure. 




N. 


S. 


E. 


w. 


1 


29.17 
47.20 
53.77 
38.92 


23.56 




17.19 
43.53 




2 


18.25 
37.37 




3 




38.66 


4 


32.06 




22.06 










Sum 


55.62 


55.62 


60.72 


60.72 







In this example the sum of the northings is equal to the 
sum of the southings, and the sum of the eastings to the sum 
of the westings; hence the work balances and no correction is 
required. 

Double Mekidiak Distance. 

16. Def. The double meridian distance of a course is douUe 
the distance of its middle point from some given meridian 
chosen at pleasure. 

Thus, if A BCD be the plot of the survey whose field-notes 
are given in the preceding section, E, F, G, H the midde points 

N 




of the sides, and N8 a meridian passing through the most west- 
ern station J, then '^KE is the double meridian distance of the 
course or distance AB, 2LFihQ double meridian distance of 
the course BC, and so on. 

Since AE \^ equal to EB, ?A{E is equal to PB] hence. 



DOUBLE MLBIDIAN DISTANCE. 



131 



The double meridian distance of the first course is equal to 
its departicre. 

Also, LF = PB -\-ZF', 

therefore 2LF = ^KF + FB + EC. 

But 2KFis the double meridian distance of the first course, 
FB its departure, and EC the departure of the second course; 
hence, 

The double meridian distance of the secofid course is equal 
to the double meridian dista7ice of the first course 2jIus its de- 
parture plus the departure of the second course. 

Again, MO = MT -^ TV -^ VG 

= FB ^ EC - SC-{- VG, 
and %MG = %KE -{- FB -\- 2EC - 2SC + SC 

= 2LF-\- EC - SC. 

But 2LF is the double meridian distance of the second course, 
EC its departure, and SC the departure of the third course; 
hence we have, in general: 

The double meridian dista^ice of any course is equal to the 
double meridia7i disfa^ice of the preceding course p)lus its de- 
parture plus the departure of the course itself. 

The double meridian distance of the last course is evidently 
equal to its departure with the opposite sign. 

This agreement affords a check in forming the column of 
double meridian distances. 

From the diagram, it is evident that the area of ABGD 
is equal to the sum of the trapezoids FBCQ^ QCDX mmViS 
the sum of the triangles ABF, AFX, and that twice the 
sum of the trapezoids minus twice the sum of the triangles 
is equal to twice the area of A BCD. 

The general form of operation is shown in the following 
table: 



Station. 


Course. 


N. Lat. 


S. Lat. 


D. M. D. 


North Areas. 
{Triangles.) 


South Areas. 
(Trapezoids.) 


1 
3 


AB 
BO 
CD 
DA 


AP 

"xa' 


"W 


2KE 
2LF 
2MG 
2NH 


2KEXAP 


2LFXPQ 
2MGXQX 


3 




4 


2JVHXAX 



132 



LAND SXIBVEYINa. 



Whence we have for finding the area the following general 

Rule : Multiply the double meridian distance of each course 
by its latitude, placing the product in the column of north 
areas when the latitude is north, and in the column of south 
areas when the latitude is south; the difference of the sums 
of the two columns divided by 2 is the area required. 

Thus, from the preceding field-notes we have the following 
results : 





Dis. 


Latitude. 


Depature. 


D. M. D. 


North 

Areas. 

{Triangles) 


South 


Sta. 


N. 
+ 


s. 


E. 

+ 


w. 


{Trape- 
zoids.) 


1 

2 


29.17 
47.20 
53.77 
38.92 


23.56 
*32.b6 


'l8.25 
37.37 


17.19 
43.53 



'38.66 
22.06 


17.19 
77.91 

82.78 
22.06 


404.9964 


1421.8575 


3 




3093.4886 


4 


707.2436 
















55.62 


55.62 


60.72 


60.72 




1112.2400 


4515.3461 



Area = 



4515.3461 - 1112.2400 
2 



= 1701.55305 square chains 

= 170 acres roods 24.8 perches. 



EXAMPLES. 



1. Find the area of the survey whose field-notes are as 
follows: 



Station. 


Bearing. 


Distance. 


1 
2 
3 
4 
5 
6 


N. 11° 37' E. 

E. 
S. 10°46'W. 
S. 28° 5' E. 
S. 62° 23' W. 
N. 29° 26' W. 


25.12 
32.06 
19.71 
30.17 
31.00 
41.04 



SUPPLYING OMISSIONS. 



133 



2. Find the bearing and length of the line joining the 
first and fourth stations. 

Ans. Bearing, K 81° 5'.4. Distance, 33.845 ch. 

3. Find the angle between the first and fourth courses. 

Ans. 39° 42'. 

Supplying Omissions, 

1 7 . When the length and bearing of a side are wanting, the 
latitude and departure of this side may be found as follows: 
Calculate the latitudes and departures of the giyen sides in 
the usual manner; then the difference between the northings 
and southings will be the northing or southing of the unknown 
side, and the difference between the eastings and westings 
will be the easting or westing of the unknown side. If we 
require the bearing and length of this side, we have in the 
right-angled triangle whose sides are the latitude Q) and de- 
parture {d) found as just explained, and whose hypothenuse 
(c) is the distance required, 

tan B ■= -j', c = I sec B. (4) 

In the same manner we may 
find the length and bearing of -^ 
a side which cannot be measured 
by reason of obstructions. Thus, 
suppose BC is an inaccessible line ^ 
whose extremities only are known, 
and C invisible from B. In the 
right-angled triangle BBC, the 
sides BB and BC are known, 
being respectively the differences 
between the northings and south- 
ings and the eastings and westings 
of the entire survey ABCB; hence 
the bearing BBC, and the distance 
BC, can be found as above. 

Irregiclar Boundaries. 

18. When one or more of the bounding lines of a survey, is 
a broken line, such as the bank of a river or other body of 




134 



LAND SUBVETINQ. 



water, a random line is run between two convenient points in 
the boundary of the survey, and perpendiculars measured 
from this line to the angles 
of the neighboring broken ^ 
line. Thus, let AB be a 
random line; measure Am, 
mn, np, pB, and the per- 
pendiculars ma, nt, pc. 
Then the areas of the tri- 
angles Amn, Bpc and the 
trapezoids mabn, nlcp can be found. 




Laying Out and Dividing Land. 

19. In the laying out as well as in the dividing of land the 
surveyor must depend for the most part on his knowledge of 
algebra, geometry and trigonometry. Only a few examples 
can be given here for the purpose of showing the student the 
general mode of procedure. 

1. Lay out 10 acres in the form of a rectangle, the differ- 
ence of whose length and breadth is 6 ch. 

Let A denote the area, x the length, y the breadth, and 
d the difference of x and y, then we have 



xy = A and 



y = d. 



Whence we get x = i Yd' + 4.A -^ id = 13.44 ch; 
y =^ i Vd' + 4^ - id = 7.44 ch. 

2. Lay out 100 acres in the form of a parallelogram one 
of whose sides shall be 50 chains and shall make with the 
other side an angle of 85° 5'. 

Put A = the area in square chains, a = 84° 5', J = 50, 
and x = the unknown side. Then we shall have 



whence 



ix sm a = A', 
A 



-z- cosec a 





20.107 chains. 



3. Lay out a farm of 52 acres in the form of a parallelo- 
gram whose sides shall be 30 and 18 chains, and the bearing 
of the former N. 53° 7' E. Determine the bearing of the 
other side. . Ans. S. 52° 31'. 6 E. 



DIVIDING LAND. 



135 



4. From a triangular field of given area cut off the nth. 
part by a line parallel to the base. 

Let ABG be the field, and DE the 
dividing line. 

Since BDEC is to be the 7^th part, 

n — 1 
ADE = .ABC, and, by the prop- 




erty of similar triangles, we have 

ABC: ADE:: AB' 
n -1 



or 



whence 



ABC: 



AD 



n 



AD\ 
ABC:: AB' 



AD' 



= ABi/- 



Having thus determined the point D, run the line DE parallel 
to the base BC. 

5. From a triangular field, the bearings and lengths of 
whose sides are given, cut off a tri- 
angle of given area by a line having 
a given bearing. 

Let ABChe the field, the bearin gs 
and lengths of whose sides, AB, 
BC, CA, are given, and ADEi\iQ a 
triangle cut off by the line DE, 
whose bearing is also given. The 
angles of the triangle ADE are q 
known from their bearings (§14). 

Let AD = X, AE = y, and area of ADE = n. 
have 

xy sin A = 2n 

X sin E 




We shall 



and 



whence we easily get 



y 



y 



sini>' 
__ . / %7i sin E 



sin A sm D^ 
. / 2n sin D 
sin A sin E' 



6. To divide a farm in the form of a trapezoid, the bear- 
ings and lengths of whose sides are given, into two parts, in 



136 



LAND SUBVETINa. 



the ratio of m : w, by a line joining the parallel sides and 
drawn from a given point in one of the sides. 

Nl B g Q o 




/ 



H P 



E 



The angle BAD is known from the bearings of the 
sides ABj AD; therefore 

BH = AB sin A, 

Put BC = a, AD = h, and BH = h, P the given pointy 
and PQ the dividing line. Divide BG and AD each into 

: n; that is, take 



two parts in the ratio of m 
ma 



Then 



Ba=:^ 



GC =z 



m -\- n 
na 



and AE — 



ABGE 



m -{- n 
hm{a -\- h) 



and ED = 



BGDE 



m -{- n 

nb 
m + n^ 
hn(a -\- h)^ 
2(m + ny 



2{m -\- n) 

which are in the ratio oi m : n. 

Bisect GE in F, join PF, and produce it to meet the 
other side in Q. Then, since the triangles PEE and FGQ 
are equal, we have 

ABQP: QGDP :: m : n. 

Levelling. 

30. Def. A level surface is one which coincides with or is 
parallel to the surface of tranquil water. 

Such a surface, therefore, partakes of the spheroidal 
form of the earth, but in all operations of levelling it is 
regarded as perfectly spherical. Points situated on such a sur- 
face are said to be on the same level, and a line drawn on it 
is called a line of true level. The difference of level, then, of 
two places is simply the difference between their distances 




LEVELLING. 137 

from the centre of the earth. A line of apparent level of a 
place is a straight lii]e tangent to the earth's surface at that 
place, and is therefore perpendicular to the 
plumb-line. Thus, if PB represent the sur- 
face of still water, as that of the ocean, then P 
and B are on the same level; and if PA be 
drawn tangent to the surface at P, then PA 
is the line of apparent level. The difference 
AB, is the correction to the apparent level for 
curvature, and is computed as follows: 

Put PC, the radius of the earth = r; 

PA, the distance = d', 

and AB, the correction = x» 

Then, by geometry, we have 

(2r + x)x = d\ 
or x^ + 2rx = d^, 

from which to determine x-, but since x is always very small 
compared with r, the term x^ may be omitted without appreci- 
able error. Therefore we have 

^ = ^-; (5) 

that is, the difference between the true and apparent level is 
very nearly equal to the square of the distance divided by the 
diameter of the earth; and since the latter is constant, the 
correction for curvature varies as the square of the distance. 

EXAMPLES. 

1. Find the correction for curvature for a distance of one 
mile, the diameter of the earth being taken at 7912 miles. 

Here we have x = -ztt.-x = 8.008 inches. 
7912 

2. What is the difference between the true and apparent 
level for half a mile ? Ans. 2 inches. 

3. An object is visible at the distance of 26 miles; find 
its height. Ans. 150.37 yards. 

4. The difference between the true and apparent level of 
two places, is 100 feet; what is the distance? 

Ans. 12.241 miles. 



138 



LAWD 8TIBYETING. 



31, The Engineer's Level is shown in the annexed en- 
graving. It consists essentially of an accurately-constructed 
level attached to the under side of a telescope which is supported 
by two brass forks, called Y's from their shape. Two semi- 
circular clamps or rings hold the telescope securely in the Y 's, 
and at the same time allow it to revolve in them. One of the 
Y 's has a nut and screw by which it can be raised or lowered, 
and the tube containing the level is furnished at each end 





Levelling Instkujient. 

with adjusting screws by means of which it may be moved in 
both vertical and horizontal directions, 

A graduated scale is attached to the upper side of the 
level-tube for the purpose of enabling us to bring the bubble 
more accurately to the centre. 

The tripod-head, levelling-screws, clamp- and tangent- 
screw are in all respects similar to those of the transit theodo- 
lite, already described. 

Adjustments. 

22. 1st. To adjust the Line of OoUimation. 
Having set up the instrument firmly on the tripod, bring, 
by means of the levelling- and tangent-screws, one of the 



THE ENOmEEB'S LEVEL. 139 

wires in the telescope into exact coincidence witli some sharply- 
defined mark — such as the intersection of two black lines on 
a white wall — at the distance of 150 or 200 yards. Turn the 
telescope in the Y's half round, so that the same wire may 
be brought to bear on the mark. 

If the wire coincides with the mark, the wire is already 
adjusted; but if not, bring it back half-way by means of the 
capstan-headed screws which move the diaphragm holding 
the wires. Eepeat the operation until the wire coincides with 
the mark in both positions of the telescope, and then proceed 
in the same manner with the other wire. 

2d. To adjust the Level-tule. 

Turn the instrument on its yertical axis until the telescope 
is over either pair of opposite levelling-screws, clamp the in- 
strument, and bring the bubble to the middle of the tube. 
Revolve the telescope- in the Y's through an angle of from 
30° to 90°; then if the bubble run to one end, correct half 
the error by the adjusting-screw, which gives a horizontal 
motion, and bring the level back to its original position. 
Again bring the bubble to the centre, revolve the telescope in 
the Y's, and repeat the operation until the bubble remains 
in the centre when the telescope is revolved in either way. 
Lastly, bring the bubble to the centre by the levelling-screws 
and reverse the telescope in the Y's. If the bubble now run 
to one end, lower that end or raise the other by the adjusting- 
screw which gives a yertical motion to the level, till half the 
correction is made. Again bring the bubble to the centre, 
and repeat the operation till the reversion can be made with- 
out producing any change in the position of the bubble. 

3d. To adjust the Y^s, or to mahe the Line of Collimation 
at Right Angles to the Vertical Axis, 

Place the telescope over either pair of opposite levelling- 
screws, and bring the bubble to the centre of the tube. Turn 
the instrument on its vertical axis half-way round; then if 
the bubble does not remain in the centre, make half the 
correction by the screw which elevates or depresses one of 
the Y's, and the other half by the levelling-screws, and repeat 
the operation over the other pair of levelling-screws until the 
bubble remains stationary in the middle of the tube during 



140 



LAND SUBYEYma. 



an entire reyolution of the instrument on its vertical 
axis. 

Use of the Level. 

23, The instrument having been accurately adjusted and 
the tripod firmly placed in the ground, bring the bubble to the 
middle of the tube, and place the wires precisely in the com- 
mon focus of the eye-piece and object-glass, as directed in the 
case of the transit theodolite; then the intersection of the 
wires will indicate the line of apparent level. 

To find the Difference of Level of Tiuo Places. 

Let A and D be the places whose diiference of level is 
required. 

Set up the instrument at some station E between any two 
convenient points A and B', take the backsight to A and 
observe the height Aa on the perpendicular staff AH, which 
is divided to inches and tenths of an inch. Set up the staff 
at B and observe the height of the foresight at h. Now 




remove the instrument to some other station, as F, and 
observe the height Be of the backsight and the height Cd of 
the foresight, and so on until the point D is reached. Then 
the difference between the sum of the backsights and the sum 
of the foresights will be the difference of level; for the sum of 
the backsights 

= Aa ^ Be -^ Ce = Aa -^ Bh ^ he ^ Ce, 
and the sum of the foresights 

^Bl-\- Gd-\-Df= Bl-\- Ge-^de-^Df 
=^ Bb ^ Ce + Mc, 
therefore sum of backsights — sum of foresights 

= Aa -{• to — Mc = — AL,the difference of level. 



TOPOGBAPHIGAL SUBVETING. 



141 



When the sum of the foresights is greater than the sum of 
the backsights, the point D is below A, which is indicated by 
the minus sign; but if the reverse is true, then the point D 
is above A, which is denoted by the plus sign. 

In this operation it is not essential that the intermediate 
stations E, F, (r be in a direct line between the two places. 

A record of the backsights and foresights is kept in the 
following form and constitutes what is called field-notes, as 
in the following example: 



Station. 


Backsights. 


Foresights. 


1 
3 

3 

4 


ft. in. 
6 8.5 
2 11.9 
2 10.2 
1 7.3 


ft. in. 

2 7.1 

4 0.6 

5 8.7 

3 8.6 


Plirn?? ,,,.-.. ^ ,.,, , 


14 1.9 
16 1.0 


16 1 






Difference of level 


= - 1 11.1 





TOPOGKAPHICAL SURVEYING. 

24. Topogi'aphical surveying has for its object the delinea- 
tion on paper of the form of the surface of a survey, the 
situation with regard to the boundaries, of ponds, streams, 
marshes and other objects. 

The surface is supposed to be intersected by horizontal 
planes at equal distances from each other, and the curves 
which are formed by the intersection of these planes with the 
surface of the ground being transferred to paper will indicate 
the variation in the slope or inclination of the ground, for it 
is evident that these curves will be nearer together or farther 
apart according as the inclination is steep or gentle. 

Thus, let A BCD be a survey divided by a stream GH, 
running through it. 

The lowest point is at H, at which place a permanent mark, 
called a hench-marh, and take the horizontal plane through 
this point for the plane of reference. Suppose the ground 



142 



LAND SUBVETINa. 



to be intersected by a horizontal plane 4 feet aboye H, in the 
undulating lines marked 4 on each side of the stream. Let 
another horizontal plane be drawn 8 feet above H and inter- 
secting the ground in the lines marked 8^ and so on till the 
summit S is reached, which we find on the right to be 32 




feet above //. On the left the ground rises more gently till it 
reaches P, 16 feet above H, and then descends again towards 
the point A, 



EXAMPLES. 

1. The magnetic bearing of an object is S. 2° 13' W. and 
its true bearing is S. 2° 57' E. Find the variation of the 
needle. Ans. K 5° 10' W. 

2. The true bearing of an object is S. 70° 5' E., and the 
variation of the needle is N. 5° 1' W. Find the magnetic 
bearing. Ans. S. 65° 4' E. 

3. The declination of a Ursse Majoris (Dubhe) is 
N. 62° 22' 38". Find its azimuth at the time of its greatest 
eastern elongation at a place in latitude 37° 15' N. 

Ans. 35°37'.6E. 

4. The declination of /? Ursge Minoris is N". 74° 38'. Find 
its azimuth at its greatest western elongation at a place in 
latitude 46° 17'. Ans. N. 22° 32'. 8 W. 



EXAMPLES. 



143 



5. Find tlie area of the survey whose field-notes are as 
follows: 



] 



Station. 


Bearing. 


Distance. 


1 
3 
3 
4 
5 


K 37° 1' E. 
S. 46° O'E. 
S. 10° 5' W. 

S. 78° 7' W. 
N. 24° 43' W. 


19.76 ch. 

14.00 

wanting 

12.50 

15.72 



Ans. 41 acres 3 roods 14.3 perches. 

6. Divide the survey whose field-notes are given in § 13, 
into two equal parts by a line parallel to the third side, 
and determine the point of intersection of the dividing line 
with the fourth side. 

7. Find the azimuth of Polaris at greatest elongation, on the 
1st of January, 1886, at places whose latitudes are 29° 54' N". 
and 50° ]^. ^ 

8. Show that Polaris may cross the meridian three times 
in a solar day, twice above the pole and once below. 

9. In what latitude is the azimuth of Polaris at its greatest 
elongation, equal to twice its polar distance? 

Ans. 59° 59'. 5 N. 



AS'SWEES. 



i 15. I. 60°; 12°; 48°; 24°; 36°; 36°; 24°; 48°; 12°; 60°. 
2. 120°. 3. III., 255°. 4. 135°; 22.5 sec. 
6. 24°; 48°; 96°; 192°. 
1 16. 8. 57.296 metres. 9. 3438 metres; 206265 metres. 
10. 143.24 feet. 11. 1.53 feet. 12. 5.4 miles. 

13. 27000 miles. 14. 614.5 feet. 15. 69.04 miles. 
16. 866000 miles. 17. 8.82 minutes. 18. 2.73 inches. 
19. 108°. 
36. I. sin y = 0.60; cos y = 0.80; tan y = 0.75; 
cot y = 1.33; sec y = 1.25; cosec y = 1.66. 
2. sin y = 0.80; cos y = 0.60; tan y = 1.33; 
cot y = 0.75; sec y — 1.66; cosec y = 1.25. 
29.16. tan = 1 -^ m; cot = m. 17. 17° 7' or 72° 53'. 

22. 24°.3. 23. ^U ± Vrn'-i] and|Uq= Vm''- 4V 

^ 24. 63°.5. 25. 60°. 32. 2. 33. ^cti. . 34. 1. 35. 1. 
36. + !• 37- 1- 38. tan''^; — cot^a;. 
39. 1. 40. 0. 44. 1.414. 45. cos i?; + sec ic. 

46. sin 60° =-ir-; cos 60° = ^; tan 60° = |/3; 

cot 60° = -V; sec 60° = 2; cosec 60° = -^. 
1^ ' V3 



. ^.o VlO + 2J^ ^., VS'-I 

47. sm 72 =^ ; cos 72 = 



4 



tan 72° = ^^- + ^; cot 72°= ^- ^ 



Vb-1 ' '1^10 -]-%Vt 

4 ,0 4 



sec 72° = ; cosec 72° = - 



V5 - 1 1^10 + 2 ^5 

[1451 



146 AN8WEM8. 

48. 51° 20'. 49- ^^° ^8'- 5o- T 1 + -^-. 

§ 33. 1. a^ 38° 40'; c = 6.40. 2. « = 58° 0'; c = 9.43. 

3. ^ ==27° 10'; c = 94.53. 4. a= 0° 11'; c = 876.6. 

5. «=:86° 6'; c =: 761.0. 6. «f=17°32'; c = 28646. 

§ 34. I. a= 51° 3'; ^ = 5.66. 2. a = 55° 46'; 6? = 13.23. 

S. a = 43° 1'; b = 88.04. 4. «: = 72° 42'; a = 9.38. 
§35.1.^ = 18.61; 5=5.81. 2. « = 3.09; 5=10.48. 

3. « = 8.575; c = 11.81. 4. ^ = 6.21; c = 6.28. 

5. ^P = 989; OP = 992. 6. 170 feet. 

7. 3671 metres = h. 

8. Base = 10.80; sides = 7.618 and 11.184. 

9. Distance of bases = 79.38 feet; height of second 

tower =64.12 feet; distances of snmits = 90.7 feet. 

11. r of circumscribed 6^=13.07 metres, and r of 

inscribed circle = 12.07 metres. 

12. 14° 56'. 13. 1110 sq. feet of tin. 14. 30.05 metres. 

15. Length of diagonals = 8.6; angles = 109° and 71°. 

16. Altitude = c Vl - cos'/?. 

17. Distance of nearer tower = 756.7 metres, and their 

common height = 165.5 metres. 

18. OX = 1.35; 19. Side = r V2; area = 2r'; 
XF = 0.74; side = V2; area = 2. 

JVP = 1.24; 20. Side = 1.175; area = 2.377. 
Qli = 1.13; 21. Side = 1.453r; area = 3.633r' 
QO = 1.25; 22. a = 53° 8'; /3 = 36° 52'; 
0X= 1.68. sides = 15 and 20. 

23. Course, 30°; time, 8°^ 39^6. 

24. Base = 2; Alt. = ^6 +2 VK 

25. 3762, 3426, 2798, 1978 and 1222 miles. 

26. Distance from base = 1000 feet, angle = 29° 2'; 

distance from summit = 1143 feet. 

27. 1.24; 1.24. 28. 2.54; 1.27; 2.20. 
29. 13, 12, 5; 22° 37', 67° 23'. 30. 30°; 60°. 
31. 33° 41', 56° 19'; 7.2, 10.8. 32. 78° 28'. 
33. 49.85 inches. 34. 2.47 feet. 35. 66.3 feet. 



«2 



ANSWERS. 



147 



7,6, 124.9 feet. 37. 11.5 and 9.64 miles, 

38. Height = 201.6 feet; distance = 379.1 feet. 



39- 

42. 

44. 
46. 
48. 
49. 

i36. 3 

i50.i4. 
16. 



46° 50'. 40. 234000 miles. 41. 88 feet. 



7.22, 4.17 and 8.33. 
12.92 feet. 45. 19' 
73° 45' and 16° 15'. 
110° and 70°. 
Sides = 8 and 6. 



43. 24.4 inches. 
28' and 70° 32'. 
47. r = 1.098. 



106° 16' and 73° 44'. 



The angles at diagonals 



51° 50'; 38° 10'. 
60°; 120°; 240° 

For 135°: 
sin = 0.707; 
cos= - 0.707; 
tan= — 1.000; 



10'. 



4. 38' 
300°. 

For 225°: 



For 315°: 



cot = 
sec = 
cosec = 
17. sines 



1.000; 
1.414; 
1.414. 



sm = 
cos = 
tan = 
cot = 
sec = 
cosec = 



- 0.707; 
-0.707; 

1.000; 
1.000; 

- 1.414: 



sm = 
cos = 
tan = 
cot =: 
sec = 



1.414. cosec 



= 0.866; 0.866; 0.500; - 0.500; 



cosines 



tan 



15. 

20. 
30- 

34. 



- 0.866 
0.500; - 

- 0.500 
= 1.732; - 1 

- 1.732 
and 300°. 19. 



- 0.500. 

0.500; - 

0.500: 



0.866; 



, 0.866. 
.732; - 0.577; 0.577; 

- 0.577. 
120° and 240°. 



0.707; 
0.707; 
1.000; 
1.000; 
1.414; 
1.414. 
0.866; 

0.866; 

1.732: 



60° 

51° 20'; 128° 40'; 231° 20'; 308° 40'. 

26° 34'; 206° 34'. 

335° 42' and 114° 18'; 155° 42' and 294° 18'. 

323°.l. 35. {i - a) sin x, 

36. {m — n) sin x. 37. 

40. 1. 41. — 2m sin a. 
43. 2(??z — 2^) sin X, 44. 
45. — 2jj sin a. 46. 

58. 1.1 = 642.9; c = 616.7. 

3. «= 132.3; 5 = 92.4. 

4. AG = 1700 metres; BG = 1793 metres. Perp 

1685 metres. 



= 0. 38. 1. 39. 1. 

42. — %n cos a. 
a secV — h co&QC^a. 
4r sec a. 

2. a=. 1.91; c= 19.75. 



148 ANSWERS. 

5. ^5 = 3.1783; ^(7^:2.3759. The perpendiculars, 

2.6555, 2.3118, 1.9850. 

6. The sides are 234.24; the altitude = 224.66 metres. 

7. 84° 19'; 95° 41'; 12.345; 20.667. 

8. 29.47, 23.31 and 33.93. 9. 25.57, 36.76 and 33.59. 
10. Area = 2601 sq. ft.; sides = 58.67 and 72.57 feet. 
12. 37.4, 94.3, 94.3 metres. 13. 37.5, 33.2,29.3 metres. 

14. 18.10, 6.34 and 6.68. 

15. The angles = 97° 48', 126° 5'; the sides = 57.27, 

24.74, 36.05 and 30.33. 

16. 14.30; 25.35. 17. 39.96; 53.16. 

18. 545.0 and 671.9. 

19. Sides = 48.18, 53.41; base = 44.94. 

§ 59. 1. 13=. 45° 52' or 134° 8'; c = 44.858 or 11.434. 

2. a= 17° 3' or 162° 57'; c = 66.556. 

3. J3= 36° 18' or 143° 42'; c = 52.71 or 5.98. 

4. a = 147° 28' or 0° 54'; a = 35.52 or 1.04. 

5. r= 29° 24' or 150° 36'; b = 15.4.2. 

6. Distances = 139.95 and 31.58. 

§63. I. 36° 52'; 53° 8'; 90°. 2. 63° 9'; 45° 9'; 71° 43'. 

3. 6° 0'; 63° 28'; 110° 32'. 

4. Diagonal, 16.3; angles, 67°, 113°. 
6. 78° 28'; 101° 32'; 57° 7'; 122° 63'. 

10. 14.7, 12.0 and 10.3 metres. 
12. 4.5 and 35.3 metres from yertex. 14. 108° 13'. 
15. 7.5. 19. 26.40 and 8.68. 20. 1893 square feet. 
21. 45°17'. 22. 120.6 feet. 25. 126.6 feet. 
26. 411 and 138 feet from A = on the road. 
28. 105° 55', 55° 18'; 1L253, 3.770. 
§63. I. 133° 56'; 11° 52'. 2. 93° 28'; 50° 39'. 
3. 80° 9'; 1°58'. 4. 46° 12'; 26° 0'. 

5. /? = 64° 27'; y = 17° 3'. a = 9966. 

6. 272.3 metres; direction = N. 22° 48' W. 

7. Diagonals = 37 and 20.7; angles = 18° 55', 34° 12', 

35° 13' and 91° 39'. 
S. a = 2.339; i = 5.004; a = 76° 24'; /? := 103° 36'. 

9. 74° 57'. 10. 67° 58', 63° 8' and 48° 54'. 



ANSWERS. 149 

11. 68°41'; lll°i9'; 52°!'; 127°59'; diagonals = 54 and 

33.1. 

12. 132.6 and 82.2 metres. 13. 123.7 and 63.8 metres. 

14. Distance = 248.6 knots; direction of northern ship 

= N. 1° 46' W. 

15. A = 57° 22'; B = 107° 15' or 35° 1'; D = 92° 29' or 

10° 31'; C = 102° 54'. 

16. ^ + ^ 17. 122° 7' and 21° 16'. 
m — n 

18. y + /•'' + ^r; jt?^ + g^ + ^2^; 5^^ + r'^ + g'r. 

20. A = 95° 7'; ^ = 82° 59'; 6^ = 76° 16'; Z> = 105° 38'. 

§64. I. a = 133°36'; /? = 30°8'; c = 52; area=546sq. metr. 

5. 125° 18'; 67° 0'; 78° 25'; 89° 17'; area 451.1. 

9. 52°46'; length =:314yards. 10. 7.43 knots E.73°25'K 

II. 440.4 feet. 12. 3.85 sq. feet. 13. 26.4 feet. 

14. 88750 sq. feet and 148830 sq. feet. 15. 6084 sq. feet. 

16. Distance = 10640 feet S. 25° 36' E. 

17. 600 feet, and meets the base in the middle. 

18. 597. 6 feet, and meets the base 495 feet from one end. 

19. 42, 58 and 78 feet; 1165, 2414 and 1725 sq. feet. 

20. Height = 68 yards. 21. 104.5 feet. 

23. 18, 24 and 30 feet. 24. 38° 10' and 51° 50'. 
§65.1. 9> = 12°12'; r = 6.139. 2. 9? = 32° 33'; r = 0.153. 

3. ^z= 87°30'.7; r = 194860. 
§66, 1. 309° 8'; 305.7. 2. 163° 41'; 9658.0. 

3. 267° 11'; 1.966. 4. 359° 28'; 396.7. 

5. 261° 29'; 18.205. 6. 101° 34'; 14.28. 

7. r = 266.5; /x ^ 871.8; d = 151° 39'; x = 98° 1'. 

8. e = 62° 51'; r = 221.4. 

§67. I. 36.071 sin(/f + 130°2'.7). 2. 8.2422 sin((9'+31°19'). 

3. ^ sin {6 + 53° 8'). 5. - 14° 26' or + 55° 32'. 

1/& «i 

6. 60° 33' or 172° 35'. 

§ 68. I. A. = 322° 20'.5; j3 = 28° 59'.2; r = 1.8366. 
2. X = 110° 0'; /? = 60°2'.4; r = 24.277. 



150 



ANSWERS. 



5- 
6. 

7. 

83. I. 

3- 

4. 

5- 

7. 

14. 

15- 

16. 

17. 

18. 

19. 

20. 



Spheeical Tkigokometry. 

a^h = 30° 22'; C= 98° 25'. 

a = l) = ^r 41'; (7 == 128° 40'. 

a = h=^ 73° 41'; (7 =148° 36'. 

^ = 31° 40'; B = 114° 36'; c = 42° 20'. 

40° 4'. 2. ^ = 81°24'.4; 5 = 34° 34'. 2. 

^ = 44° 17' or 135° 43'; 

^ = 98°ll'.4or261°48'.6; 5= 101°46'.3. 

& = 6° 53'.5 or 173° 6'.5; c = 32° 11'.9 or 147°48'.l. 

c = 81°27'.3; b = 76° 50'.5; a = 49° 15'.6. 

a = 90°; b = 30°; A = 90°; B = 30°. 

Angles at perpendiculars = 35° and 55°; the per- 
pendicular = 52° 25'. 

1,1 1 

cos a — -— =; cos = --=; cos c = 77. 

COS a = — 1£-; cos c> = y — ; cos c = ~T7~. 
1/3 2 To 

COS a = 1; cos 5 = — 1; cos c = — 1. 

cos a = — 1; COS 5 = — 1; cos c = 1. 

1-3 VS 



cos ^ = 



2V3 



=; cot ^ = — ; cos c = -— -. 



sm 5 = ?r; sm i/ = —7^; 
3 ^^3 



sm C = — ^nr 



f3 



83. 


I. 


^=:i?=:52°21'; (7 =62° 50'. 




2. 


c = 57°3'; B = 122° 56'. 


85. 


I. 


106°38'.0; 82° 4'.4; 49° 49'.2. 




2. 


148° 56'; 143° 40'; 149° 58'. 




3. 


19° 29'; 30° 33'; 130° 58'. 




4- 


158° 30'; 58° 14'; 61° 30. 




5. 


22° 34'; 39° 54'; 125° 24'. 


86. 


I. 


76° 30'; 93° 18'; 122° 8'. 




2. 


a = 59° 27'.4; b = 242° 53'.8 or 117° 6'.2; 
c = 123° 20' or 236° 40'. 




3. 


a = 79° 6'.0; b = 102° 6'.7; c = 89° 13'.5. 




4. 


155° 22'; 75° 13'; 92° 16'. , 



AN8WEB8. 151 

5. 95° 52'; 85° 52'; 75° 52'. 

6. 23° 37'; 42° 2'; 58° 19'. lo. The sides=109° 28'. 
II. Ig cos « = 9.8557; Ig cos I = 9.8557?^; Ig cos c = 

9.8941^. 
§8T. I. B= 28°49'.0; (7= 109° 44'.6; « = 104° 28'.0. 

2.^= 56°37'.3; ^ = 132°12'.9; c= 20° 21'.2. 

3. A= 42° 40'; B = 17° 34'; c = 63° 45'. 
§88.1. C= 94°15'.8; a= 27° 1'.4; J = 49° 24'.0. 

2. A= 116° 15'.1; ^ = 162° 25'.9; c = 137° 20'.1. 

3. a =151° 11'; i= 70° 16'; C= 75° 32'. 

4. ^=123° 21'; I)= 47° 49'; (7 =159° 38'. 
§ 89. I. 120° 31' and 70° 20'. 2. 99° 41' and 64° 23'. 

3. 95° 33' and 92° 39'. 4. 79° 10' and 80° 19'. 

§93.1. (7= 75° 17'; .4 = 97° 10'; a = 95° 56'. ■ 

2. ^ = 25° 42'; A = 149° 32'; a = 95° 48'. 

3. i = 75° 11'; C = 61° 31'; c = 92° 18'. 

4. c =75° 54'; A= 84° 8'; a= 86° 3'. 

I. 69° 13'. 2. The angles = 70° 32' and 60°. 

3. 90°, 36° 52' and 53° 8'; 180°, 0° and 0°. 

. 4. Inclination = 39° 11'; 

Triangle = 63° 26', 63° 26' and 53° 8' 

5. Latit. = -19° 28'. 6. K 49°20'E.; 3675 miles. 

7. 83° 27' = 5772 miles. 8. 939 miles from New York. 
9. Latitude = - 49° 21' in longitude 25° 34' East from 

Greenwich. 



Logarithms of Numbees. 



[158] 



logj^eithms of numbees. 



65. The logarithm of a number consists of an integer and 
a decimal fraction. 

The decimal fraction is called the mantissa of the loga- 
rithm. 

The integer is called the characteristic of the logarithm. 
A table of logarithms gives only the mantissag. 

66. To find the mantissa of the logarithm of a given 
number. 

Case I. When the number has three or fewer figures. 

EuLE. Find in the left-hand column that line luhich con- 
tains the first tiuo figures of the number, and select that column 
which has the third figure at its top. The four figures in this 
column on the line selected form the mantissa of the logarithm. 

Examples. Mantissa of log 6 = .7782; 
^Mogl70 = .2304; 
"log 707 = .8494. 

Case II. When the number has four or more figures. 

EuLE. Find the mantissa of the first three figures as in 
Case I., and call this mantissa m. 

Tahe the difference D between this mantissa and that next 
following it in the table. 

Imagine the third figure of the given number to be followed 
by a decimal point, and multiply D by the fourth and fifth 
figures considered as decimals. 

Add the product to m; the sum will be the logarithm required. 

Example. To find mantissa of log 17762. 
mant. of log 177 = .2480 
D = 24; .62 X 24 = 15 

mant. log 17762 = .2495 
Note. Tlie product of D by the fourth and fifth figures may be 
found from the table of proportional parts. Find D in the first column 
and the fourth and fifth figures on top. The student should be able to 
find the product and add it in his head. 



156 LOGARITHMS. 

6*7. Characteristics, The characteristic of a logarithm 
is less by unity than the number of figures preceding the 
decimal point in the number to which it corresponds. 

Examples, The mantissa last found gives the following 
logarithms: 

log 17762 = 4.2495: 
log 1776.2 = 3.2495] 
log 177.62 = 2.2495; 
log 1.7762 = 0.2495. 
Every time we move the decimal point one place toward 
the left we divide the number by 10 and diminish the loga- 
rithm by unity, which leaves the mantissa unchanged. 

If the number is a decimal fraction, the characteristic is 

— 1 when the figure following the decimal point is differ- 

ent from zero; 

— 2 when one zero follows the decimal point; 

— 3 when two zeros follow it, and so on. 

In these cases the minus sign is written adove the char- 
acteristic to show that it belongs to the characteristic alone 
and not to the mantissa. 

Examples, log 0.17762 =1.2495; 
log 0.017762 =2.2495; 
log 0.0017762 = 3.2495; 
etc. etc. 

EXERCISES. 

Find the logarithms of the following numbers: 



1. 


701. 


6. 


7032. 


11. 


1.0246. 


2. 


700. 


7. 


70.32. 


12. 


2.0324. 


3. 


70. 


8. 


0.7032. 


13. 


51.623. 


4. 


7. 


9. 


0.007032. 


14. 


11.111. 


5. 


0.7. 


10. 


2956. 


15. 


3.243. 



QS. To find the number corresponding to a given loga- 
rithm. 

EuLE. 1. Find in the talle, if possible, the mantissa of 
the given logarithm. The corresponding figures in the left- 
hand column and at the top of the column are the figures of 
the required numher. 



LOGARITHMS OF NUMBERS. I57 

2. If the mantissa is not found in the table, find the next 
smaller mantissa. The three corresponding figures are the 
first three figures of the required number. 

3. Take the excess of the given mantissa over the next 
smaller one in the table and divide it by the difference between 
the two consecutive mantissce in the table. 

4. The result is a decimal fraction to be written after the 
three figures already found, the decimal poi7it being dropped. 

5. Having thus found the figures of the number, insert the 
decimal point corresponding to the given characteristic accord- 
ing to the rule 0/ § 67. 

Example 1. Find the number whose logarithm is 4.7267. 

We find in the tables that to the mantissa .7267 corre- 
spond the figures 533. The characteristic being 4, there must 
be five figures before the decimal point, so that we have 

Number = 53300. 

Ex. 2. Find the number whose logarithm is 1.2491. 
"We find from the tables 

Next smaller mantissa, 2480; iV^ = 178. 
Excess of given mantisssa, 11. 

Difference of mantissae in table = 2504 - 2480 = 24. 

Figures to be added to iV^ = -J^ = .46. We may perform 
this division in the head by the table of proportional parts, 
by entering the table in line 24 and taking out the successive 
figures of the quotient from the top of the table. 

Therefore the figures of the number are 17846. The char- 
acteristic being 1, there are two figures before the decimal 
point, so that 

Number = 17.846. 

EXERCISES. 

Find the numbers corresponding to the following loga- 
rithms: 

1. 1.2032. 6. 2.0212. 

2. 0.7916. ' 6. 4.0343. 

3. 0.0212. 7. 3.1922. 

4. T.0212. 8. 2.8282. 



158 LOGARITHMS. 

69, Change of Negative Characteristics to Positive Ones, 
We may avoid the use of negative characteristics by increas- 
ing them by 10, and then indicating a subtraction of 10 from 
the logarithm. For example, instead of writing 

log 0.0017762 = 3.2195, 
we may write 

log 0.0017762 = 7.2195 - 10. 

We may also, in most cases, omit writing — 10 after the 
logarithm, leaving it to be understood. The characteristics 
will then be given by the following rule: 

The characteristic of the logarithm of a decimal fraction is 
9 minus the numher of zeros foUoiDing the decimal ])oint. 
Examples. log 0.8624 =9.9357; 

log 0.08624 = 8.9357; 
log 0.008624 = 7.9357; 
the expression — 10 being omitted. 

EXERCISES. 

Find the logarithms of : 

1. 35. 5. 92.25. 9. 0.10562. 

2. 0.35. 6. 0.9225. 10. 1.0042. 

3. 0.0035. 7. 0.73262. 11. 3.0035. 

4. 0.00035. 8. 0.15143. 12. 1.0007. 

'^O. Use of Logarithms. The following properties of 
logarithms are demonstrated in treatises on algebra: 

I. The logarithm of a product is equal to the sum of the 
logarithms of its factors. 

II. The logarithm of a quotient is found hy subtracting 
the logarithm of the divisor from that of the dividend. 

III. The logarithm of any poicer of a number is equal to 
the logarithm of the number multiplied by the exponent of the 
poiver. 

IV. The logarithm of the root of a number is equal to the 
logarithm of the number divided hy the index of the root. 

We thus derive the following rule for finding a joroduct: 
To find the product of several factors by logarithms. 
EuLE. Add the logarithms of the several factors. Enter 



LOGARITHMS OF NUMBEBS. 159 

the table tvith the sum as a new logarithm^ and find the num- 
ber corresponding to it. 

This number is the product required. 
Example 1. To find the continued product 2x6x8. 
log 2, 0.3010 
log 6, 0.7782 
log 8, 0.9031 

Sum of logs, 1.9823 = log product. 
The number corresponding to this logarithm is found to 
be 96, which is the product required. 

Ex. 2. To find the quotient of 147 -^ 21. 
log 147, 2.1673 
log 21, 1.3222 

Difference, 0.8451 
We find this difference to be the logarithm of 7, which is 
the required quotient. 

Ex. 3. To find the quotient arising from dividing the con- 
tinued product 98 X 102 X 148 by the continued product 
21 X 37 X 68. 

log 21, 1.3222 log 98,1.9912 

log 37, 1.5682 log 102, 2.0086 

log 68, 1.8325 log 148, 2.1703 

Sum = log divisor, 4.7229 Sum = log dividend, 6.1701 

log divisor, 4. 7229 

Difference = log quotient, 1.4472 
Looking into the table, we find the number corresponding 
to this logarithm to be 28, which is the required quotient. 

Note. The student will notice that we have found this quotient 
without actually determining either the divisor or dividend, having 
used only their logarithms. If he will solve the problem arithmetically, 
he will see how much shorter is the logarithmic process. 

Ex. 4. To find the cube of 15. 
We have log 15 = 1.1761 

3 



3.5283 = log 3375 
Hence 3375 is the required power. 



160 LOGARITHMS. 

Ex, 5. To find the cube root of 125. 
3 I 2.0969 



0.6990 
The index of the root being 3, we divide the logarithm of 
125 by it. Looking in the tables, we find the number to be 5, 
which is the root required. 

EXERCISES. 

Compute the following products, quotients, powers and 
roots by logarithms. 

1. 12^ Ans. 1738. 5. 5i#^-. Ans. 31. 

-|f)3 91/1^2 

2. ^. Ans. 48. 6. ^. Ans. 1.4457. 

„ 3.9'. 91.78 . ,„„ „ 0.5296'' . „ „„,„ 
3- -l3^3I:3-• ^'''- 1^8- ^- o:9303-- ^'''- ^-^^^S. 

4 ^£. Ans. 138. 8. '-^^§^i A. 0.3888. 

|/121 0.05632 

't'l. The Arithmetical Complement. Let it be required to 

find the value of the fraction r-^; that is, the reciprocal of 

l./v5 

1.25. By the rule we must subtract log 1.25 from log 1; that 
is, from 0. Since this will give a remainder with a negative 
characteristic, which we shall want to increase by 10, the 
easiest course will be to increase log 1 by 10 before we sub- 
tract, which will give a remainder increased by 10. The work 
will then be : 

log 1 = 10.0000 - 10 

log 1.25 = 0.0969 

log (1 : 1.25) = 9.9031 - 10 
Hence 

i= '■'■ 

Def. The arithmetical complement of a logarithm 
is the remainder obtained by subtracting the logarithm from 
or 10. 
The arithmetical complement of the logarithm of a number 



TBIGONOMETBIG FUNCTIONS. 161 

is called the co-logarithm of the number, and is written 
co-log. 

The co-logarithm may be obtained by subtracting each 
figure of the logarithm, beginning at the left, from 9, except 
the last significant figure, which must be subtracted from 10. 

72. The logarithm of a quotient may be obtained by add- 
ing the co-logarithm of the divisor to the logarithm of the 
dividend. 

1655 
Example. To compute the value of . 

log 1655 = 3.2188 
co-log 1.25 = 9.9031 

log quotient = 3.1219 
quotient = 1324 

EXERCISES. 

Find the values of the following fractions: 

111 

1. ~. 4. — -— . 7. 



35 0.9915 0.0125 

2.^. 5. 1 " ^ 



1.027 0.125 0.001569* 

1 ^57 ^ 1.0022 



• 1.0055* • 0.08252* * 0.9867* 

Logaritlims of Trigonometric Functions. 

73. By means of the table following that of logarithms 
we may find the logarithms of the trigonometric functions of 
any angle. The tables give the logarithms of the functions 
because these are commonly more convenient to use than the 
functions themselves. 

The method of using the table will depend upon the 
magnitude of the angle. 

74:. Case I. Angles less than 45°. In this case we 
find the number of degrees and minutes in the left-hand 
column, and then, in the four columns of large figures follow- 
ing, the log sine, log tangent, log cotangent and log cosine, 
according to the headings at the top of the column. 



162 LOGABITHMS. 

The interpolation to single minutes or tenths of minutes 
is then performed in the same way as in the case of loga- 
rithms of numbers. To facilitate the interpolation, tables of 
proportional parts are given in the margin after the first five 
degrees of the table. 

Example. To find the logarithm of the trigonometric 
functions of 8° 16'.3. 

Entering the table with the next smaller angle, 8° 10', we 
find the values of the four functions given in the first line 
below together with the differences following them. Then to 
interpolate for the rest of the angle, 6'. 3, we must take 0.63 
of 87. The difference 87 not being found in the margin, we 
notice what difference it falls between and interpolate accord- 
ingly. We thus find that corresponding to 6' is the difference 
52, and corresponding to 0'.3 the difference 3. These num- 
bers are therefore to be added to the sine 8° 10' in order to 
obtain the sine 8° 16'. 3. 

For the tangent the difference is 89, which is given in the 
marginal table. To 6' corresponds 53.4, and to 0'.3 corre- 
ponds 2.7. 

For the cotangent the differences are the same as for the 
tangent except that they are negative. In the case of the 
cosine the difference is so small that the interpolation may be 
made in the head. The following results are then obtained: 

L. sin. L. tan. L. cot. L. cos. 

8° 10' ....9.1525 9.1569 0.8431 9.9956 



+ 52 +53 - 53 - 1 

+ 3 +3 - 3 



p.p. for 6' . 
0'.3. 

For 8° 16'.3. . . .9.1580 9.1625 0.8375 9.9955 

We ma3? obtain the same quantities by entering the table 
with 8° 20' and interpolating to — 3'. 7, which is the quantity 
by which the given angle differs from 8° 20'. The work will 
then be as follows: * .- ', ^ 





L. sin. 


L. tan. 


L. cot. 


"£. cos. 


For 8° 20' . . 


..9.1612 


9.1658 


0.8342 


9.9954 


- 3' .. 


.. -26 


- 27 


+ 27 


+ 1 


- 0'.7.. 


.. - 6 

..9.1580 


- 6 
9.1625 


+ 6 
0.8375 





For 8° 16'.3.. 


9.9955 



TBIGONOMETBIG FUNGTI0N8. 163 

The careful computer should interpolate by both methods 
and see if the results agree. He should also make sure that 
the number he gets falls between the proper ones in the 
table. 

To find the secant and cosecant we use the theorem of §26. 
Because the secant is the reciprocal of the cosine, we have 



Hence 



log secant = co-log cosine; 
log cosecant = co-log sine. 

log CSC 8° 16'. 3 = 0.8420; 
log sec 8° 16'.3 = 0.0045. 

EXERCISES. 

Find the logarithms of all six trigonometric functions of 
the following angles: 

1. 26° 38'.4. 4. 0° 59'.2. 7. 21° 5'.6. 

2.44° 0'.9. 5. 5°39M. 8. 32° 4'. 3. 

3. 4°16'.3. 6. 22° 10'. 7. 9. 40° 8'. 2. 

75. Case II. Angles letiveen 45° and 90°. Because 
the function of any angle is the complementary function 
of the complement of the angle (§ 21), we may use the same 
table for the angle and its complement by reyersing the 
meaning of the function. The complementary angle is written 
at the bottom and in the right-hand column of the table, and 
the names of the complementary functions are written at the 
bottom. Hence, to find the functions of angles between 45° 
and 90°, we haye the rule: 

Fi7id the angle at the bottom and right of the talle and 
the name of the function at the iottom. 

The number in the corresponding column, ivhen interpolated 
in the tisual way, will give the required logarithm of the func- 
tion. 

As the first example, and to show the relations of the two 
complementary angles, we seek the functions of the comple- 
ment of the angle already used, namely, 81° 43'. 7. 



164 


LOGARITHMS. 








L. sin. 


L. tan. 


L. cot. 


L. COS. 


81° 40' .. 


..9.9954 


0.8342 


9.1658 


9.1612 


DiS. for 3' . . 


.. +1 


+ 27 


- 27 


-26 


'' '' 0'.7.. 





+ 6 


- 6 


- 6 



81° 43'.7.... 9.9955 0.8375 9.1625 9.1580 

L. sec 81° 43'. 7 = 0.8420; 
L. CSC 81° 43'. 7 = 0.0045. 

EXERCISES. 

Find the six functions of the following angles: 

1. 63° 21'. 6. 4. 89° 0'.8. -7. 89° 22'. 5. 

2. 45° 59'.1. 5. 84° 36'.2. 8. 53° 20'.3. 

3. 85° 43'. 7. 6. 85° 1'.5. 9. 62° 9'.2. 

•^G. Oa3E III. Angles in the second quadrant, or le- 
tween 90° and 180°. 

In this case the functions are found by the formulaB of 
§ 48, which give the following rule: 

Subtract ^^"^ from the angle, and enter the table with the 
remainder. 

Tahe out the function cornplementary to that required, and 
hy giving the proper sign ive shall have the required function 
of the angle. 

Examples. sin 107° 29' = cos 17° 29'; 

tan 107° 29' = - cot 17° 29'; 

cot 107° 29' = - tan 17° 29'; 

cos 107° 29' = - sin 17° 29'; 

etc. etc. 

EXERCISES. 

Find the trigonometric functions of the following angles : 

1. 116°38'.4. 4. 153°21'.6. 7. 179° 22'.5. 

2. 134° 0'.9. 5. 135° 59'.1. 8. 143° 20'.3. 

3. 94° 16'.3. 6. 175° 43'. 7. 9. 152° 9'.2. 

'3"7. Case IV. When the angle is between 180° and 270°. 

Rule. Subtract 180° from the angle. Enter the table 
with the remainder and tahe out the required function. 

By assigning the proper algebraic sign, we shall have the 
required function of the angle. 



TRIGONOMETRIC FUNCTIONS. 165 



Examples, sin 197° 25' 
cos 252° 35' 
tan 249° 6' 


= - sin 17° 25'; 
= - cos 72° 35'; 
= + tan 69° 6'. 






EXERCISES. 




Compute 


the yalues of the following expressions 
a - 92.62, 
I — 0.9253, 
§ = 73° 29', 
y^ 131° 3': 

1. a co^ f3 -{- i sin {^ + y). 

2. b sin /3 — a cos (/? + y). 


when 



3. ai cos {y — /3) — — tan (/ + /5). 

•rS. Case V. Angles hetween 270° «7?.^? 360°. 
EuLE. Subtract 270° /ro?/i tlie angle, enter the talle with 
the remainder and take out the functions complementary to 
those required. 

By assigning the proper algebraic signs toe shall have the 
required fu7ictions of the given angle. 

Examples, sin 302° 42' -- - cos 32° 42'; 
tan 343° 17' = - cot 73° 17'; 
sec 321° 49' = esc 51° 49'. 

EXERCISES. 

Find the logarithms of the six trigonometric functions of: 
1. 302° 42'. 2. 327° 18'. 

I' 9. Method of luriting the Algebraic Signs. In the pre- 
ceding examples we have written before the logarithms the 
sign which really belongs to the number and not to the loga- 
rithm. In practice this is the most convenient method for 
the computer to follow, but it must always be remembered 
that the algebraic sign as thus written does not imply that 
the logarithm is to be added or subtracted accordingly. The 
addition or subtraction of the logarithm must always depend 
upon whether the number to which the logarithm belongs is 
a factor or divisor, and the sign of the result must theji be 
determined by the rules of algebra. 



THREE- AISTD FOUR-PLACE 
TABLES 



OF 



LOGARITHMS AND TRIGONOMETRIC 

FUNCTIONS. 



[167] 



168 



TABLE OF FOUR-PLACE LOGABITHMS. 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


oooo 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 

12 
13 
14 
15 
16 

17 

18 
19 


0414 
0792 
II39 

I46I 
I76I 
2041 

2304 

2553 
2788 


0453 
0828 

1173 
1492 
1790 
2068 

2330 

2577 
2810 


0492 
0864 

I205 

1523 
I8I8 
2095 

2355 
2601 

2833 


0531 
0899 

1239 

1553 
1847 
2122 

2380 
2625 
2856 


0569 

0934 
1271 

1584 

1875 
2148 

2405 

2648 
2878 


0607 
0969 
1303 
1614 
1903 
2175 
2430 
2672 
2900 


0645 
1004 
1335 
1644 

1931 
2201 

2455 
2695 

2923 


0682 
1038 
1367 

1673 

1959 
2227 

2480 
2718 
2945 


0719 
1072 
1399 
1703 
1987 
2253 
2504 
2742 
2967 


0755 
1 106 
1430 

1732 
2014 
2279 

2529 
2765 
2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 

22 
23 
24 
25 

26 

27 
28 
20 


3222 

3424 
3617 

3802 

3979 
4150 

4314 
4472 
4624 


3243 
3444 
3636 

3820 

3997 
4166 

4330 

4487 
4639 


3263 
3464 
3655 

3838 
4014 
4183 

4346 
4502 
4654 


3284 
3483 
3674 
3856 
4031 
4200 

4362 
4518 
4669 


3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 


3324 

3522 

3711 
3892 
4065 
4232 

4393 
4548 
4698 


3345 
3541 
3729 

3909 

4082 
4249 

4409 
4564 
4713 


3365 
3560 

3747 

3927 
4099 
4265 

4425 
4579 
4728 


3385 
3579 
3766 

3945 
4116 

4281 

4440 
4594 
4742 


3404 
3598 
3784 
3962 
4133 
4298 

4456 
4609 
4757 


30 


4771 


4786 


4800 


4814 


4829 


i 4843 


4857 


4871 


4886 


4900 


31 
32 
33 
34 
35 
36 

37 
38 
39 


4914 
5051 
5185 

5315 
5441 
5563 
5682 
5798- 
5911 


4928 
5065 
5198 
5328 
5453 
5575 

5694 
5809 
5922 


4942 
5079 
5211 

5340 
5465 

5587 

5705 
5821 

5933 


4955 
5092 
5224 

5353 
5478 
5599 

5717 
5832 
5944 


4969 
5105 
5237 
5366 

5490 
5611 

5729 
5843 
5955 


4983 
5119 
5250 

5378 
5502 
5623 

1 5740 
i 5855 
1 5966 


4997 
5132 
5263 

5391 
5514 
5635 

5752 
5866 
5977 


5011 

5145 
5276 

5403 
5527 
5647 

5763 

5877 
5988 


5024 

5159 
5289 

5416 
5539 
5658 

5775 
5888 

5999 


5038 
5172 
5302 

5428 
5551 
5670 

5786 
5899 
6010 


40 


6021 


6031 


6042 


6053 


6064 


i 6075 


6085 


6096 


6107 


6117 


41 

42 
43 
44 
45 
46 

47 

48 
40 


6128 
6232 
6335 

6435 
6532 
6628 

6721 
6812 
6902 


6138 
6243 
6345 
6444 
6542 
6637 

6730 
6821 
6911 


6149 
6253 
6355 

6454 
6551 
6646 

6739 
6830 
6920 


6160 
6263 
6365 
6464 
6561 
6656 

6749 
6839 
6928 


6170 
6274 
6375 
6474 

6571 
6665 

6758 
6848 
6937 


6180 
6284 
6385 
6484 
6580 
6675 
6767 
6857 
6946 


6191 
6294 
6395 

6493 
6590 

6684 

6776 
6866 
6955 


6201 
6304 
6405 

6503 
6599 
6693 

6785 
6875 
6964 


6212 
6314 
6415 

6513 
6609 
6702 

6794 

6884 
6972 


6222 

6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 


50 


6990 


6998 


7007 


7016 


7024 


1 7033 


7042 


7050 


7059 


7067 


51 
52 
53 
54 


7076 
7160 
7243 
7324 


7084 
7168 
7251 
7332 


7093 
7177 

7259 

7340 


7101 

7185 
7267 

73^8 


7110 
7193 

7275 
7356 


7118 
7202 
7284 

7364 


7126 
7210 
7292 

7372 


7135 
7218 
7300 

7380 


7143 
7226 
7308 

7388 


7152 

7235 
7316 

7396 



TBIGONOMETBIG FUNCTIONS. 

Pkopoktional Paets. 



169 



D 


1 


2 


3 


4 


5 


6 


7 


8 


9 


43 


4-3 


8.6 


1.2.9 


17.2 


21.5 


25.8 


30.1 


34-4 


38.7 


42 


4.2 


8.4 


12.6 


16.8 


21.0 


25.2 


29.4 


33.6 


37.8 


41 


4.1 


8.2 


12.3 


16.4 


20.5 


24.6 


28.7 


32.8 


36.9 


40 


4.0 


8.0 


12.0 


16.0 


20.0 


24.0 


28.0 


32.0 


36.0 


39 


3-9 


7.8 


II. 7 


15.6 


19-5 


23.4 


27.3 


31.2 


35.1 


38 


3.8 


7.6 


II. 4 


15-2 


19.0 


22.8 


26.6 


30.4 


34.2 


37 


3-7 


7.4 


II. I 


14.8 


18.5 


22.2 


25.9 


29.6 


33.3 


36 


3.6 


7.2 


10.8 


14.4 


18.0 


21.6 


25.2 


28.8 


32.4 


35 


3-5 


7.0 


10.5 


14.0 


17.5 


21.0 


24.5 


28.0 


31.5 


34 


3-4 


6.8 


10.2 


13.6 


17.0 


20.4 


23.8 


27.2 


30.6 


33 


3-3 


6.6 


9.9 


13.2 


16.5 


19.8 


23.1 


26.4 


29.7 


32 


3-2 


6.4 


9.6 


12.8 


16.0 


19.2 


22.4 


25.6 


28.8 


31 


3-1 


6.2 


9.3 


12.4 


15.5 


18.6 


21.7 


24.8 


27.9 


30 


3-0 


6.0 


9.0 


12.0 


15.0 


18.0 


21 .0 


24.0 


27.0 


29 


2-9 


5.8 


8.7 


II. 6 


14.5 


17.4 


20.3 


23.2 


26.1 


28 


2.8 


5.6 


8.4 


11. 2 


14.0 


16.8 


19.6 


22.4 


25.2 


27 


2.7 


5.4 


8.1 


10.8 


13.5 


16.2 


18.9 


21.6 


24-3 


26 


2.6 


5.2 


7.8 


10.4 


, 13.0 


15.6 


18.2 


20.8 


23.4 


25 


2.5 


5.0 


7-5 


10. 


12.5 


15.0 


17.5 


20.0 


22.5 


24 


2.4 


4.8 


7.2 


9.6 


12.0 


14.4 


16.8 


19.2 


21.6 


23 


2.3 


4.6 


6.9 


9.2 


II. 5 


13.8 


16. 1 


18.4 


20.7 


22 


2.2 


4.4 


6.6 


8.8 


II. 


13.2 


15.4 


17.6 


19.8 


21 


2.1 


4.2 


6.3 


8.4 


10.5 


12.6 


14.7 


16.8 


18.9 


20 


2.0 


4.0 


6.0 


8.0 


10. 


12.0 


14.0 


16.0 


18.0 


19 


1.9 


3.8 


5-7 


7.6 


9.5 


II. 4 


13.3 


15.2 


17. 1 


18 


1.8 


3.6 


5.4 


7.2 


9.0 


10.8 


12.6 


14.4 


16.2 


17 


1.7 


3.4 


5.1 


6.8 


8.5 


10.2 


II. 9 


13.6 


15-3 


16 


1.6 


3-2 


4.8 


6.4 


8.0 


9.6 


II. 2 


12.8 


14.4 


15 


1.5 


3.0 


4.5 


6.0 


7.5 


9.0 


10.5 


12.0 


13-5 


\i 


1.4 


2.8 


4.2 


5.6 


7.0 


8.4 


9.8 


II. 2 


12.6 


13 


1.3 


2.6 


3-9 


5.2 


6.5 


7.8 


9.1 


10.4 


II. 7 


12 


1.2 


2.4 


3.6 


4.8 


6.0 


7.2 


8.4 


9.6 


10.8 


11 


I.I 


2.2 


3.3 


4.4 


5-5 


6.6 


7.7 


8.8 


9.9 


10 


I.O 


2 


3.0 


4.0 


5.0 


6.0 


7.0 


8.0 


9.0 


9 


0.9 


1.8 


2.7 


3-6 


4-5 


5.4 


6.3 


72 


8.1 


8 


0.8 


1.6 


2.4 


3.2 


4.0 


4.8 


5.6 


6.4 


7.2 


7 


0.7 


1.4 


2.1 


2.8 


3.5 


4.2 


4.9 


5.6 


6.3 


6 


0.6 


1.2 


1.8 


2.4 


3.0 


3.6 


4.2 


4.8 


5.4 


5 


0.5 


1.0 


1.5 


2.0 


2.5 


3.0 


3.5 


4.0 


4-5 


4 


0.4 


0.8 


1.2 


1.6 


2.0 


2.4 


2.8 


3.2 


3.6 



170 



L0OARITHM8. 
Peoportional Parts. 



D 


1 


2 


3 


4 


5 


6 


7 


S 


9 


8 


0.8 


1.6 


2.4 


3.2 


4.0 


4.8 


5.6 


6.4 


7.2 


7 


0.7 


1.4 


2.1 


2.8 


3-5 


4.2 


4.9 


5.6 


6.3 


6 


0.6 


1.2 


1.8 


2.4 


3-0 


3-6 


4.2 


4.8 


5-4 


5 


0.5 


I.O 


1.5 


2.0 


2.5 


3-0 


3-5 


4.0 


4-5 


4 


0.4 


0.8 


1.2 


1.6 


2.0 


2.4 


2.8 


3-2 


3.6 



The table on the preceding page shows the product of 
each difference between two con seen tiye logarithms by each of 
the nine digits. 

As an example of its use let it be required to find the loga- 
rithm of 1934. 

We see from the table that log 193 = 2856, and that the 
difference between this and the next logarithm is 22. We find 
22 in column D/and opposite it, in column 4, the number 
8.8. The nearest integer to 8.8 is 9, which we add to 2856, 
and find 3.2865 for the required logarithm of 1934. 

As a second example find log 10. 648. 

For 106 we find .0253 with the difference 41. 

Opposite D = 41 and under 4 we find 16.4. Under 8 we 
find 32.8, one tenth of which is to be added to 16.4, making 
19.68. The nearest integer is 20, which being added to .0253 
giTes 1.0273 for the required logarithm. 

Ex. 3. log 22606 = 4.3552. 

Ex. 4. To find the number of which the logarithm is 
1.0097. 

The mantissa next smaller is .0086 corresponding to the 
number 102. The difference is 42 and the excess of the given 
logarithm is 11. Looking in line 42 we find the product next 
smaller than 11 to be 8.4 in column 2. The excess of 11 over 
8.4 is 2.6. Multiplying by 10 gives 26; the nearest number to 
which is in column 6. Hence the required number is 10.226. 

Note. A practical computer will find it advantageous to cut this 
leaf out, paste it on the right-hand margin of the table, and fold it in 
when not in use. He will then have the whole table before him at one 
opening, and have no leaves to turn over in taking out logarithms. 



TABLE OF FOUR-PLACE LOGARITHMS. 



171 



l\o. 





1 1 


2 


3 


! 4 


5 


6 


7 


8 


9 


55 
56 


7404 
7482 


7412 
7490 


7419 
7497 


7427 
7505 


7435 
7513 


7443 
7520 


7451 

7528 


7459 
7536 


7466 
7543 


7474 

7551 


57 
58 
59 


7559 
7634 
7709 


7566 
7642 
7716 


7574 
7649 

7723 


7582 
7657 
7731 


7589 
7664 
7738 


7597 
7672 

i 7745 


7604 
7679 

7752 


7612 
7686 
7760 


7619 
7694 
7767 


7627 
7701 

7774 


80 


7782 


7789 


7796 1 7803 


7810 


1 7818 


7S25 


7832 


7839 


7846 


61 
62 
63 


7853 
7924 

7993 


7860 

7931 
8000 


786S 
7938 
8007 


7875 
7945 
8014 


7882 
7952 
8021 


7889 

7959 
8028 


7896 
7966 
8035 


7903 
7973 
8041 


7910 
7980 
8048 


7917 
7987 
8055 


64 
65 
66 


8062 
8129 
8195 


8069 
8136 
8202 


8075 
8142 
8209 


8082 

8149 
8215 


8089 
8156 
8222 


8096 
8162 
8228 


8102 
8169 
8235 


8109 
8176 
8241 


8116 
8182 
8248 


8122 
8189 
8254 


67 

68 
69 


8261 

8325 
8388 


8267 
8331 
8395 


8274 
8338 
8401 


8280 

8344 
8407 


8287 
8351 
8414 


8293 

8357 
8420 


8299 

8363 
8426 


8306 
8370 
8432 


8312 
8376 
8439 


8319 

8382 

8445 


70 


8451 


8457 


8463 : 8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 
72 
73 


8513 
8573 
8633 


8519 
8579 
8639 


8525 
8585 
8645 


8531 
8591 
8651 


8537 
8597 
8657 


8543 
8603 
8663 


8549 
8609 
8669 


8555 
8615 

8675 


8561 
8621 
8681 


8567 
8627 
8686 


74 
75 

76 


8692 

8751 
8808 


869S 
8756 
8814 


8704 
8762 
8820 


8710 
8768 
8825 


8716 

8774 
8831 


8722 
8779 
8837 


8727 

8785 
8842 


8733 
8791 

8848 


8739 
8797 

8854 


8745 
8802 

8859 


77 
78 
79 


8865 
8921 
8976 


8871 

8927 
8982 


8S76 
8932 
8987 


8882 
8938 
8993 


8887 
8943 
8998 


8893 
8949 
9004 


8899 

8954 
9009 


8904 
8960 
9015 


8910 
8965 
9020 


8915 
8971 
9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 
82 
83 


9085 
9138 
9191 


9090 

9143 
9196 


9096 
9149 
9201 


9101 

9154 
9206 


9106 

9159 
9212 


9112 
9165 
9217 


9117 
9170 
9222 


9122 
9175 
9227 


9128 
9180 
9232 


9133 
9186 
9238 


84 
85 
88 


9243 
9294 

9345 


9248 
9299 
9350 


9253 
9304 
9355 


9258 

9309 
9360 


9263 
9315 
9365 


9269 

9320 

1 9370 


9274 
9325 
9375 


9279 
9330 
9380 


9284 
9335 
9385 


9289 
9340 
9390 


87 
88 
89 


9395 
9445 
c>494 


9400 
9450 
9499 


9405 
9455 
9504 


9410 
9460 
9509 


9415 
9465 
9513 


: 9420 

9469 

! 9518 


9425 
9474 
9523 


9430 
9479 
9528 


9435 
9484 
9533 


9440 
9489 
9538 


90 


9lU2 


9547 


9552 


9557 


9562 


i 9566 


9571 


9576 


9581 


9586 


91 
92 
93 


9590 
9638 
968:, 


9595 
9643 
9689 


9600 
9647 
9694 


9605 
9652 
9699 


9609 
9657 
9703 


1 9614 

I 9661 

9708 


9619 
9666 
9713 


9624 
9671 
9717 


9628 

9675 
9722 


9633 
9680 
9727 


94 
95 
96 


9731 
9777 
9823 


9736 
9782 
9827 


9741 
9786 
9832 


9745 
9791 
9836 


9750 
9795 
9841 


9754 

! 9800 
' 9845 


9759 
9805 

9850 


97^3 
9809 

9854 


9768 
9814 
9859 


9773 
9818 
9863 


97 
98 
99 


9868 
9912 
9956 


9872 
9917 
9961 


9877 
9921 

9965 


9881 
9926 
9969 


9886 
9930 
9974 


9890 

9934 
9978 


9894 
9939 
9983 


9899 
9943 
9987 


9903 
9948 
9991 


9908 
9952 
9996 









00- 


0° 50'. 










L. Sin. 


Diff. 


L. Tangr. 


Diff. 


L. Cotg. 


L. Cos. 




0° 


00 




— 00 




CO 


0.0000 


O'OO" 


1 


6.4637 


3011 


6.4637 


301 1 


3-5363 


. 0000 


59 89 


2 


6.7648 


1760 


6.7648 


1760 


3-2352 


. 0000 


58 


3 


6 . 9408 


1250 


6.9408 


1250 


3-0592 


. OUUO 


57 


4 


7.0658 


969 


7-0658 


969 


2.9342 


0.0000 


56 


5 


7.1627 


792 
669 


7.1627 


792 
669 


2.8373 


. 0000 


55 


6 


7.2419 


7.2419 


2.7581 


0.0000 


54 


7 


7.3088 


580 


7-3088 


580 


2.6912 


. 0000 


53 


8 


7.3668 


512 


7.3668 


512 


2.6332 


. 0000 


52 


9 


7.4180 


457 
414 

378 


7.4180 


457 
414 

378 


2.5820 


. 0000 


51 


10 


7-4637 


7-4637 


2.5363 


0.0000 


50 89 


11 


7-5051 


7-5051 


2.4949 


0.0000 


49 


12 


7-5429 


348 


7-5429 


348 


2.4571 


0.0000 


48 


13 


7.5777 


■322 


7-5777 


322 
299 
280 
264 
248 


2.4223 


0.0000 


47 


14 


7.6099 


0^^ 

200 


7.6099 


2.3901 


. 0000 


46 


15 
16 


7.6398 


280 
264 


7-6398 


2.3602 


. 0000 


45 
44 


7.6678 


7.6678 


2.3322 


. 0000 


17 


7.6942 


248 


7.6942 


2.3058 


. 0000 


43 


18 


7.7190 


23s 


7.7190 


235 


2.2810 


0.0000 


42 


19 


7.7425 


223 
211 
202 


7-7425 




2-2575 


0.0000 


41 


20 

21 


7.7648 


7.7648 


223 
212 
202 


2.2352 


0.0000 


40 89 

39 


7.7859 


7.7860 


2.2140 


0.0000 


22 


7.8061 


194 
184 


7.8062 




2.1938 


0.0000 


38 


23 


7.8255 


7.8255 


193 
184 
178 


2.1745 


0.0000 


37 


24 


7.8439 


178 


7.8439 


2.1561 


0.0000 


36 


25 


7.8617 


7.8617 


2.1383 


. 0000 


35 


26 


7.8787 


170 

164 


7.8787 


170 
164 


2.I213 


0.0000 


34 


27 


7.8951 


158 


7.8951 


158 


2 . 1049 


0.0000 


33 


28 


7.9109 


147 

143 

138 
133 
130 
126 


7.9109 


152 
148 
142 
138 


2.0891 


0.0000 


33 


29 


7.926T 


7.9261 


2.0739 


0.0000 


31 


30 

31 


7.9408 


7.9409 


2.0591 


. 0000 


30 89 

29 


7-9551 


7.9551 


2.0449 


0.0000 


32 


7.9689 


7.9689 


2.O31I 


0.0000 


28 


33 


7.9822 


7.9823 


^34 


2.0177 


. 0000 


27 


34 


7-9952 


7.9952 


129 
126 


2.0048 


. 0000 


26 


35 
36 


8.0078 


122 


8.0078 


122 


1.9922 


0.0000 


25 
24 


8 . 0200 


8.0200 


1.9800 


. 0000 


37 


8.0319 


119 
116 


8.0319 


119 
116 


I. 9681 


. 0000 


23 


38 


8.0435 




8.0435 




1.9565 


0.0000 


22 


39 


8.0548 


113 


8.0548 


113 


1.9452 


0.0000 


21 


40 

41 


8.0658 


107 


8.0658 


110 
107 


1.9342 


0.0000 


20 89 

19 


8.0765 


8.0765 


1-9235 


. 0000 


42 


8.0870 


105 


8.0870 


105 


I. 9130 


0.0000 


18 


43 


8.0972 


100 


8.0972 


102 


1.9028 


0.0000 


17 


44 


8.1072 




8.1072 


98 
95 


1.8928 


. 0000 


16 


45 
46 


8.1169 


97 
96 


8.1170 


1.8830 


. 0000 


15 
14 


8.1265 


8.1265 


1.8735 


0.0000 


47 


8.1358 


93 


8.1359 


94 


I. 8641 


. 0000 


13 


48 


8.1450 


92 
89 
88 


8.1450 


91 


1.8550 


0.0000 


12 


49 


8.1539 


8.1540 


90 
87 


I . 8460 


0.0000 


11 on 


50 


8.1627 


8.1627 


1.8373 


. 0000 


10 89 


1 L. Cos. 


Diff. 


L. Cotg. 


Diff. 


L. TanpT. 


L. Sin. 





89° 10' - 90° 0' 



172 









0° 50' - 


1° 40'. 










L. Sin. 


Diflf. 


L. Tang. 


Diflf. 


L. Cotg. 


L. Cos. 




0°50' 


8.1627 


86 


8.1627 


86 


1-8373 


0.0000 


10' 89° 


51 


8.1713 


84 
83 


8.1713 


85 
82 


1.8287 


o\ 0000 


9 


52 


8.1797 


8.1798 


1.8202 


0.0000 


8 


53 


8.1880 


8.1880 


I. 8120 


9.9999 


7 


54 


8.1961 


81 
80 


8.1962 


82 


1.8038 


9.9999 


6 


55 

56 


8.2041 


78 


8 . 2041 


79 

79 
76 
76 
74. 


1.7959 


9.9999 


5 

4 


8.2119 


8.2120 


1.7880 


9.9999 


57 


8.2196 


77 


8.2196 


I . 7804 


9.9999 


3 


58 


8.2271 


75 


8.2272 


1.7728 


9.9999 


2 


59 


8.2346 


75 


8 . 2346 




1-7654 


9.9999 


1 




1 


8.2419 


73 
71 


8.2419 


10 
72 
71 


I. 7581 


9.9999 


89 
59 


8.2490 


8.2491 


1-7509 


9.9999 


3 


8.2561 


71 

69 
69 
67 
66 
66 


8.2562 


6q 


1-7438 


9.9999 


58 


3 


8.2630 


8.2631 


6q 


1.7369 


9.9999 


57 


4 


8 . 2699 


8.2700 


uy 
67 

66 
66 


1 . 7300. 


9.9999 


56 


5 
6 


8.2766 


8.2767 


1.7233 


9.9999 


55 
54 


8.2832 


8.2833 


I. 7167 


9.9999 


7 


8.2898 


64 
63 
63 
62 
60 


8.2899 


64 
63 
63 
61 
61 


I.7101 


9.9999 


53 


8 


8 . 2962 


8.2963 


1.7037 


9.9999 


52 


9 


8.3025 


8.3026 


1.6974 


9.9999 


51 


1 10 

11 


8.3088 


8.3089 


I.69II 


9.9999 


50 88 
49 


8.3150 


8.3150 


1.6850 


9.9999 


12 


8.3210 


60 


8.3211 


60 


1.6789 


9-9999 


48 


13 


8.3270 




8.3271 




1.6729 


9.9999 


47 


14 


8.3329 


59 


8.3330 


59 


1.6670 


9.9999 


46 


15 
16 


8.3388 


59 
57 


8.3389 


59 
57 


I.6611 


9.9999 


45 
44 


8.3445 


8.3446 


1.6554 


9.9999 


17 


8.3502 


57 

56 


8.3503 


57 

56 


1.6497 


9.9999 


43 


18 


8.3558 


8-3559 


I. 6441 


9.9999 


42 


19 


8.3613 


55 


8.3614 


55 


1.6386 


9.9999 


41 


1 20 

21 


8.3668 


55 
54 


8.3669 


55 
54 


I. 6331 


9.9999 


40 88 
39 


8.3722 


8.3723 


1.6277 


9.9999 


22 


8.3775 


53 


8.3776 


53 


1.6224 


9.9999 


38 


23 


8.3828 


53 


8.3829 


53 


I.6171 


9.9999 


37 


24 


8.3880 


52 


8.3881 


52 


I.6119 


9.9999 


36 


25 
26 


8.3931 


SI 
51 


8.3932 


5^ 
51 


1.6068 


9-9999 


35 
34 


8.3982 


8.3983 


I. 6017 


9.9999 


27 


8.4032 


50 


8.4033 


50 


1.5967 


9.9999 


33 


28 


8.4082 


50 


8.4083 


50 


I. 5917 


9.9999 


32 


29 


8.4131 


49 


8.4132 


49 


1.5868 


9.9999 


31 


1 30 

31 


8.4179 


48 
48 


8.4181 


49 

48 


I. 5819 


9.9999 


30 88 
29 


8.4227 


8.4229 


I. 5771 


9.9998 


32 


8.4275 


48 


8.4276 


47 


1-5724 


9.9998 


28 


33 


8.4322 


47 


8.4323 


47 


1-5677 


9.9998 


27 


34 


8.4368 


46 


8.4370 


47 


1.5630 


9.9998 


26 


35 
36 


8.4414 


46 
45 


8.4416 


46 
45 


1-5584 


9.9998 


25 
24 


8.4459 


8.4461 


1-5539 


9.9998 


37 


8.4504 


45 


8.4506 


45 


I - 5494 


9.9998 


23 


38 


8.4549 


45 


8.4551 


45 


I . 5449 


9.9998 


22 


39 


8.4593 


44 


8.4595 


44 


1-5405 


9.9998 


21 


1 40 


8.4637 


44 


8.4638 


43 


1-5362 


9.9998 


2088° 




L. Cos. 


Diff. i 


LTC^t^ 


Diff. 


L. Tang-. 


L. Sin. 





88° 20' — 89° 10' 



173 







1° 40' — 


2° 30'. 










L. Sin. 


Diff. 


L. Tang-. 


Diff. 


L. Cotg. 


L. Cos. 




1°40' 


8.4637 




8.4638 




1.5362 


9.9998 


20' 88° 


41 


8.4680 


43 


8.4682 


44 


1.5318 


9.9998 


19 


42 


8.4723 


43 


8.4725 


43 


1.5275 


9.9998 


18 


43 


8.4765 


42 


8.4767 


42 


1.5233 


9.9998 


17 


44 


8.4807 


42 


8.4809 


42 


I.5191 


9.9998 


16 


45 
46 


8.4848 


41 
42 


8.4851 


42 
41 


I. 5149 


9.9998 


15 
14 


8.4890 


8.4892 


I. 5108 


9.9998 


47 


8.4930 


40 


8.4933 


41 


I . 5067 


9.9998 


13 


48 


8.4971 


41 


8.4973 


40 


1.5027 


9.9998 


12 


49 


8.5011 


40 


8.5013 


40 


1.4987 


9.9998 


11 


1 50 

51 


8.5050 


39 
40 


8.5053 


40 
39 


1.4947 


9.9998 


10 88 
9 


8 . 5090 


8.5092 


1.4908 


9.9998 


52 


8.5129 


39 
38 


8.513I 


39 


1.4869 


9.9998 


8 


53 


8.5167 


8.5170 


39 
38 
38 
37 
38 


1.4830 


9.9998 


7 


54 


8.5206 


39 


8.5208 


1.4792 


9.9998 


6 


55 
56 


8.5243 


37 
38 


8.5246 


1.4754 


9.9998 


5 

4 


8.5281 


8.5283 


I. 4717 


9.9998 


57 


8.5318 


37 


8.5321 


1.4679 


9.9997 


3 


58 


8.5355 


37 


8.5358 


37 
36 


1.4642 


9.9997 


2 


59 


8.5392 


37 
36 
36 
36 


8.5394 


1.4606 


9.9997 


1 


2 

1 


8.5428 


8-5431 


37 
36 
36 


1.4569 


9.9997 


88 
59 


8.5464 


8.5467 


1-4533 


9.9997 


2 


8.5500 


8-5503 


1.4497 


9.9997 


58 


3 


8.5535 


35 
36 


8.5538 


35 


1.4462 


9.9997 


^l 


4 


8.5571 


8.5573 


35 


1.4427 


9.9997 


56 


5 
6 


8.5605 


34 
35 


8.5608 


35 
35 


1.4392 


9.9997 


55 
54 


8 . 5640 


8.5643 


1.4357 


9.9997 


7 


8.5674 


34 


8.5677 


34 


1.4323 


9.9997 


53 


8 


8.5708 


34 


8.5711 


34 


1.4289 


9.9997 


52 


9 


8.5742 


34 


8.5745 


34 


1.4255 


9.9997 


51 


2 10 
11 


8.5776 


34 
33 


8.5779 


34 
33 


I. 4221 


9.9997 


50 87 
49 


8.5809 


8.5812 


I. 4188 


9.9997 


12 


8.5842 


33 


8.5845 


33 


1.4155 


9.9997 


48 


13 


8.5875 


33 


8.5878 


33 


I. 4122 


9.9997 


47 


14 


8.5907 


32 


8.5911 


33 


1.4089 


9.9997 


46 


15 
16 


8.5939 


32 
33 


8.5943 


32 
32 


1-4057 


9.9997 


45 
44 


8.5972 


8.5975 


1.4025 


9.9997 


17 


8 . 6003 


31 


8 . 6007 


32 


1.3993 


9.9997 


43 


18 


8.6035 


32 


8.6038 


31 


1.3962 


9.9996 


42 


19 


8.6066 


31 


8 . 6070 


32 


1.3930 


9.9996 


41 


2 20 
21 


8.6097 


31 
31 


8.6101 


31 
31 


1.3899 


9.9996 


40 87 
39 


8.6128 


8.6132 


1.3868 


9.9996 


22 


8.6159 


31 


8.6163 


31 


1.3837 


9.9996 


38 


23 


8.6189 


30 


8.6193 


30 


1.3807 


9.9996 


37 


24 


8.6220 


31 


8.6223 


30 


1-3777 


9.9996 


36 


25 
26 


8.6250 


30 
29 


8.6254 


31 
29 


1.3746 


9.9996 


35 
34 


8.6279 


8.6283 


I. 3717 


9.9996 


27 


8.6309 


30 


8.6313 


30 


1.3687 


9.9996 


33 


28 


8.6339 


30 


8.6343 


30 


1-3657 


9.9996 


32 


29 


8.6368 


29 


8.6372 


29 


1.3628 


9.9996 


^^oryo 


2 30 


8.6397 


29 


8 . 6401 


29 


1.3599 


9.9996 


30 87° 




L. Cos. 


Diff. 


L. Cotg. 


Diff. 


L. Tang. 


L. Sin. 





87° 30' — 88° 20' 



174 











2° 


30- 


3° 20'. 








1 


L. Sin. 


Diff. 


1 L. Tang. 


1 Diff. 1 L 


. Cotg. 


L. Cos. ! 1 


2° 30' 


8.6397 




8 


.6401 




•3599 


9.9996 


30' 87° 


31 


8 


.6426 


29 
28 


8 


.6430 


29 


•3570 


9.9996 


29 


32 


8 


.6454 




8 


• 6459 


29 _ 


• 3541 


9.9996 


28 


33 


8 


.6483 


29 


8 


.6487 


2<5 


•3513 


9.9996 


27 


34 


8 


6511 


20 
28 
28 
28 


8 


• 6515 




•3485 


9.9996 


26 


35 
36 


8 


6539 


8 


.6544 


29 J 


•3456 


9.9996 


25 
24 


8 


6567 


T 


• 6571 


27 ~ 

28 


•3429 


9.9996 


37 


8 


6595 




8 


• 6599 


28 


.3401 


9-9995 


23 


38 


8 


6622 


27 
28 


8 


.6627 




.3373 


9-9995 


22 


39 


8 


6650 


8 


.6654 


27 

28 


•3346 


9-9995 


21 


2 40 

41 


8 


6677 


27 
27 


8 


.6682 


•3318 


9-9995 


20 87 
19 


8 


6704 


T 


5709 


27 ~ 


.3291 


9-9995 


42 


8 


6731 


27 


8 


6736 


27 ^ 
26 


.3264 


9-9995 


18 


43 


8 


6758 


27 
26 


8 


6762 


27 \ 

26 


.3238 


9-9995 


17 


44 


8 


6784 


26 


8 


6789 


.3211 


9-9995 


16 


45 
46 


8 


6810 


26 


8 


6815 




3185 


9-9995 


15 
14 


8 


6S37 


"8" 


6842 


27 — 

26 


3158 


9-9995 


47 


8 


6863 


26 


8 


6868 


26 


3132 


9-9995 


13 


48 


8 


6889 




8 


6894 


oA 


3106 


9-9995 


12 


49 


8 


6914 


25 
26 


8 


6920 


20 


3080 


9-9995 


11 


2 50 


8 


6940 


8 


6945 


25 J 


3055 


9-9995 


10 87 


51 


8 


6965 


25 
26 


"8" 


6971 


26 


3029 


9-9995 


9 


52 


8 


6991 




8 


6996 


25 ^ 


3004 


9-9995 


8 


53 


8 


7016 


25 


8 


7021 


25 ^ 


2979 


9-9994 


7 


54 


8 


7041 


25 


8 


7046 


25 J. 


2954 


9-9994 


6 


55 

56 


8 


7066 


25 
24 
25 


8 


7071 


25 j- 


2929 


9-9994 


5 

4 


8 


7090 


8 


7096 


25 Y 


2904 


9-9994 


57 


8 


7115 


8 


7121 


^^ I 


2879 


9-9994 


3 


58 


8 


7140 


25 


8 


7145 


2855 


9-9994 


2 


59 


8 


7164 


24 


8 


7170 


25 J 


2830 


9-9994 


1 


3 
1 


8 


7188 


24 
24 


8 


7194 


24 J 


2806 


9.9994 


87 
59 


8 


7212 


y 


7218 


24 Y 


2782 


9.9994 


2 


8 


7236 


24 


8 


7242 


24 J 


2758 


9.9994 


58 


3 


8 


7260 


24 


8 


7266 


24 J 


2734 


9-9994 


57 


4 


8 


7283 


23 


8 


7290 


24 J. 


2710 


9-9994 


56 


5 
6 


8 


7307 


24 
23 


8 


7313 


23 J 


2687 


9.9994 


55 

54 


8 


7330 


8 


7337 


24 Y 


2663 


9-9994 


7 


8 


7354 


24 


8 


7360 


23 J 


2640 


9-9994 


53 


8 


8 


7377 


23 


8 


7383 


23 J 


2617 


9.9994 


52 


9 


8 


7400 


23 


8 


7406 


23 J 


2594 


9-9993 


51 


3 10 

11 


8 


7423 


23 
22 


8 


7429 


23 J 


2571 


9-9993 


50 86 
49 


8 


7445 


8 


7452 


23 Y 


2548 


9-9993 


12 


8 


7468 


23 


8 


7475 


23 j 


2525 


9-9993 


48 


13 


8 


7491 


23 


8 


7497 


I 


2503 


9-9993 


47 


14 


8 


7513 


22 


8 


7520 


23 J 


2480 


9-9993 


46 


15 
16 


8 


7535 


22 
22 


8 


7542 


22 


2458 


9-9993 


45 

44 


8 


7557 


8 


7565 


23 Y 


2435 


9-9993 


17 


8 


7580 


23 


8 


7587 


22 


2413 


9-9993 


43 


18 


8 


7602 


22 


8 


7609 


22 


2391 


9-9993 


42 


19 


8 


7623 


21 


8 


7631 


22 


2369 


9-9993 


^^Ono 


3 20 


8 


7645 


22 


8 


7652 


21 


2348 


9-9993 


4086° 




L. Cos. 


Diff. 


L. 


Cotg. 


Diff. L. 


Tang. L. Sin. 





86° 40' - 87° 30' 



175 











3° 


20 — 


4° 10'. 












L. Sin. 


Diff. 


L. Tang. 


Diff. 


L. Cotg. 


_L 


-. Cos. 




3° 20' 


8.7645 


22 


8 


.7652 




1.2348 


9 


•9993 


40' 86° 


31 


8 


7667 




8 


.7674 


22 


1,2326 


9 


•9993 


39 


22 


8 


7688 


21 
22 


8 


.7696 


22 


I . 2304 


9 


•9992 


38 


23 


8 


7710 




8 


7717 


21 


1.2283 


9 


9992 


37 


24 


8 


7731 


21 


8 


7739 


22 


I. 2261 


9 


9992 


36 


25 
26 


8 


7752 


21 


8 


7760 


21 


I . 2240 


9_ 


9992 


35 
34 


8 


7773 


8 


7781 


1. 2219 


9 


9992 


27 


8 


7794 




8 


7802 




I. 2198 


9 


9992 


33 


28 


8 


7815 


21 


8 


7823 


21 


I. 2177 


9 


9992 


32 


29 


8 


7836 


21 


8 


7844 


21 


I. 2156 


9 


9992 


31 


3 30 

31 


8 


7857 


21 
20 


8 


7865 


21 
21 


I-2I35 


_9. 


9992 


30 86 
29 


8 


7877 


8 


7886 


1.2114 


9 


9992 


32 


8 


7898 


21 


8 


7906 


20 


I . 2094 


9 


9992 


28 


33 


8 


7918 




8 


7927 


21 


1.2073 


9 


9992 


27 


34 


8 


7939 


21 


8 


7947 




1.2053 


9 


9992 


26 


35 
36 


8 


7959 


20 
20 


8 


7967 


20 
21 


I . 2033 


9 


9992 


25 
24 


8 


7979 


8 


7988 


I. 201 2 


9 


9991 


37 


8 


7999 




8 


8008 




I. 1992 


9 


9991 


23 


38 


8 


8019 


20 


8 


8028 


20 


I. 1972 


9 


9991 


22 


39 


8 


8039 


20 


8 


8048 


20 


I. 1952 


9 


9991 


21 


3 40 

41 


8 


8059 


20 

.9 


8 


8067 


19 
20 


I -1933 


_9_ 


9991 


20 86 
19 


8 


8078 


8 


8087 


1.1913 


9 


9991 


42 


8 


8098 


20 


8 


8107 


20 


I. 1893 


9 


9991 


18 


43 


8 


8117 


19 


8 


8126 


19 


I. 1874 


9 


9991 


17 


44 


8 


8137 


20 


8 


8146 


20 


I. 1854 


9 


9991 


16 


45 
46 


8 


8156 


19 
19 


8 


8165 


19 
20 


I. 1835 


_9_ 


9991 


15 
14 


8 


8175 


8 


8185 


1.1815 


9 


9991 


47 


8 


8194 


19 


8 


8204 


19 


I. 1796 


9 


9991 


13 


48 


8 


8213 


19 


8 


8223 


19 


I. 1777 


9 


9990 


12 


49 


8 


8232 


19 


8 


8242 


19 


I. 1758 


9 


9990 


11 


3 50 

51 


8 


8251 


19 
19 


8 


8261 


19 
19 


I .1739 


9_ 


9990 


10 86 
9 


8 


8270 


8 


8280 


I. 1720 


9 


9990 


52 


8 


8289 


19 
x8 


8 


8299 


19 
18 


r .1701 


9 


9990 


8 


53 


8 


8307 


8 


8317 


I. 1683 


9 


9990 


7 


54 


8 


8326 


X9 


8 


8336 


19 


I . 1664 


9 


9990 


6 


55 
56 


8 


8345 


19 
18 
18 


8 


8355 


19 
18 


I. 1645 


9_ 


9990 


5 

4 


8 


8363 


8 


8373 


I. 1627 


9 


9990 


57 


8 


8381 


8 


8392 


19 
18 
18 


I . 1608 


9 


9990 


3 


58 


8 


8400 


19 
18 


8 


8410 


I. 1590 


9 


9990 


2 


59 


8 


8418 


8 


8428 


I. 1572 


9 


9989 


1 


4 
1 


8 


8436 


18 
18 
18 
18 
18 


8 


8446 


18 

19 
18 
18 


I. 1554 


9_ 


9989 


86 
59 


8 


8454 


8 


8465 


I-I535 


9 


9989 


2 


8 


8472 


8 


8483 


1.1517 


9 


9989 


58 


3 


8 


8490 


8 


8501 


I . 1499 


9 


9989 


57 


4 


8 


8508 


8 


8518 


17 


I. 1482 


9 


9989 


56 


5 
6 


8 


8525 


17 
18 


8 


8536 


18 
18 
18 


I . 1464 


9_ 


9989 


55 
54 


8 


8543 


8 


8554 


I . 1446 


9 


9989 


7 


8 


8560 


17 
18 


8 


8572 


I. 1428 


9 


9989 


53 


8 


8 


8578 


8 


8589 


17 


1.1411 


9 


9989 


52 


9 


8 


8595 


17 


8 


8607 


18 


I. 1393 


9 


9989 


^^ OK 


4 10 


8 


8613 


18 


8 


8624 


'' 


I. 1376 


9 


9989 


50 85 




L. Cos. 


Diff. 


L 


Cotg. 


Diff. 


L. Tang. 


L. Sin. 





85° 50' — 86° 40' 



176 











4° 


10- 


5° 0'. 










L. Sin. 


Diflf. 


L. Tan,^. 


DifiE. 


L. Cotg. ! L. Cos. I 1 


4° 10 


8.8613 




8 


8624 


18 


I. 1376 ! 9.9989 


50' 85° 


11 


8 


8630 


17 


8 


8642 




1-1358 


9.9988 


49 


12 


8 


8647 


17 

18 


8 


8659 


17 


I.1341 


9.9988 


48 


13 


8 


8665 




8 


8676 


17 
18 


I. 1324 


9.9988 


47 


14 


8 


8682 


17 


8 


8694 


I . 1306 


9.9988 


46 


15 
16 


8 


8699 


17 
17 


8 


8711 


17 
17 


7". 1 289 


9.9988 


45 
44 


8 


8716 


8 


8728 


1. 1272 


9.9988 


17 


8 


8733 


17 


8 


8745 


17 


I. 1255 


9.9988 


43 


18 


8 


8749 


ID 


8 


8762 


17 

16 


I. 1238 


9.9988 


42 


19 


8 


8766 


17 


8 


8778 


I. 1222 


9.9988 


41 


4 20 

21 


8 


8783 


17 
16 


8 


8795 


17 
17 


I. 1205 


9.9988 


40 85 
39 


8 


8799 


8 


8812 


I.I188 


9.9987 


22 


8 


8816 


17 


8 


8829 


17 
16 


I.II7I 


9.9987 


38 


23 


8 


8833 


17 


8 


8845 


I-II55 


9.9987 


37 


24 


8 


8849 


10 
16 

17 
16 


8 


8862 


17 
16 

17 
16 


1-1138 


9.9987 


36 


25 

26 


8 


8865 


8 


8878 


1.1122 


9.9987 


35 
34 


T. 


8882 


8 


8895 


1.1105 


9.9987 


27 


8. 


8898 


16 


8 


8911 


16 


I. 1089 


9-9987 


33 


28 


8. 


8914 


16 
16 
16 
16 


8 


8927 




I. 1073 


9.9987 


32 


29 


8 


8930 


8 


8944 


17 
16 

16 
16 
16 
16 
16 
16 


I. 1056 


9.9987 


31 


4c 30 

31 


8 


8946 


8 


8960 


I . 1040 


9.9987 


30 85 
29 


T. 


8962 


8 


8976 


I . 1024 


9.9986 


32 


8 


8978 


16 
16 
16 
16 


8 


8992 


I. 1008 


9.9986 


28 


33 


8. 


8994 


8 


9008 


1.0992 


9.9986 


27 


34 


8. 


9010 


8 


9024 


1.0976 


9.9986 


26 


35 
36 


8. 


9026 


8 


9040 


1.0960 


9.9986 


25 
24 


8. 


9042 


8 


9056 


1.0944 


9.9986 


37 


8. 


9057 


15 


8 


9071 


15 
16 


1.0929 


9.9986 


23 


38 


8 


9073 


ID 
16 


8 


9087 


16 


1. 0913 \ 


9.9986 


22 


39 


8 


9089 


8 


9103 




1.0897 


9.9986 


21 


4 40 

41 


8 


9104 


15 
16 


8 


9118 


15 
16 


1.0882 


9.9986 


20 85 
19 


8 


9119 


8 


9134 


1.0866 


9.9985 


42 


8 


9135 


8 


9150 


ID 


1.0850 


9.9985 


18 


43 


8 


9150 


15 
18 


8 


9165 


IS 


1.0835 


9.9985 


17 


44 


8 


9166 


8 


9180 


15 
16 

15 


1.0820 


9.9985 


16 


45 
46 


8 


9181 


15 

IS 


8 


9196 


I . 0S04 


9.9985 


15 
14 


8 


9196 


8 


9211 


1.0789 


9.9985 


47 


8 


9211 


15 


8 


9226 


15 


1.0774 


9.9985 


13 


48 


8 


9226 


15 


8 


9241 


15 


1.0759 


9.9985 


12 


49 


8 


9241 


15 


8 


9256 


15 
16 

IS 


1.0744 


9.9985 


11 


4 50 

51 


8 


9256 


15 
15 


8 


9272 


1.0728 


9.9985 


10 85 
9 


8 


9271 


T 


9287 


I. 0713 


9.9984 


52 


8 


9286 


IS 


8 


9302 


IS 


1.0698 


9-9984 


8 


53 


8 


9301 


14 


8 


9316 


14 


1.0684 


9.9984 


7 


54 


8 


9315 


14 


8 


9331 


15 


1.0669 


9.9984 


6 


55 

56 


8 


9330 


15 
15 


8 


9346 


15 

IS 


1.0654 


9.9984 


5 

4 


8 


9345 


~S 


9361 


1.0639 


9.9984 


57 


8 


•9359 


14 


8 


•9376 


IS 


I .0624 


9.9984 


3 


58 


8 


•9374 


15 


8 


.9390 


14 


I. 0610 


9.9984 


2 


59 


8 


.9388 


14 


8 


•9405 


IS 


1-0595 


9.9984 


lo^ 


5 


|8 


•9403 


IS 


8 


.9420 


IS 


1.0580 


9-9983 


o85 




1 L. Cos. 


DifE. 


T 


Cotg. 


1 Diff. 


L.Tang. 


L. Sin. 





85° 0' - 85° 50' 



177 









5° 


-13°. 






1 L. Sin. 


Dlff. 


L. Tang. 


Diff. 


L. Cotg. 


L. Cos. 




5^0'! 8.9403 


142 


8 . 9420 


143 
138 


1.0580 


9.9983 


0'85° 


10 


8.9545 


8.9563 


1.0437 


9.9982 


50 


20 


8.9682 


137 


8.9701 


1.0299 


9.9981 


40 


30 


8.9816 


M4 


8.9836 


Mb 


T.OI64 


9.9980 


30 


40 


8.9945 




8 . 9966 


M-J 


1.0034 


9-9979 


20 


50 


9 . 0070 


122 
119 
115 
113 


9.0093 


123 
120 


0.9907 


9-9977 


10 


6 


9.0192 


9.0216 


0.9784 


9.9976 


84 


10 


9-0311 


9.0336 


. 9664 


9-9975 


50 


20 


9 . 0426 


9-0453 




0.9547 


9-9973 


40 


30 


9-0539 


9.0567 




0.9433 


9.9972 


30 


40 


9 . 0648 




9.0678 


T^Q 


0.9322 


9.9971 


20 


50 


9-0755 


104 
102 
99 
97 
95 


9.0786 


105 
104 


0.9214 


9-9969 


10 


7 


9-0859 


9.0891 


0.9109 


9.9968 


83 


10 


9.0961 


9.0995 


0.9005 


9 . 9966 


50 


20 


9.1060 


9 . 1096 


98 


0.8904 


9-9964 


40 


30 


9-II57 


9.1194 


0.8806 


9-9963 


30 


40 


9-1252 


9.1291 




0.8709 


9.9961 


20 


50 


9-T345 


93 
91 
89 
87 


9-1385 


94 
93 
91 
89 
87 
R6 


0.8615 


9-9959 


10 


8 


9.1436 


9.1478 


0.8522 


9.9958 


82 


10 


9-1525 


9.1569 


0.8431 


9.9956 


50 


20 


9.1612 


85 


9-1658 


0.8342 


9-9954 


40 


30 


9.1697 


84 


9-1745 


0.8255 


9-9952 


30 


40 


9.1781 




9.1831 


84 
82 
81 
80 


0.8169 


9.9950 


20 


50 


9-i863 


80 

79 
78 


9-1915 


0.8085 


9.9948 


10 


9 

10 


9-1943 
9.2022 


9.1997 


. 8003 


9 • 9946 


81 


9.2078 


0.7922 


9.9944 


50 


20 


9.2100 


9.2158 




0.7842 


9-9942 


40 


30 


9.2176 


76 


9.2236 


78 


0.7764 


9.9940 


30 


40 


9-2251 


75 


9-2313 


77 


0.7687 


9.9938 


20 


50 


9.2324 


73 
73 
71 


9-2389 


70 
74 
73 


O.761I 


9-9936 


10 


1.0 


9.2397 


9.2463 


0.7537 


9-9934 


80 


10 


9.2468 


9-2536 


0.7464 


9-9931 


50 


20 


9-2538 


70 
68 
68 


9 . 2609 


'16 


0.7391 


9-9929 


40 


30 


9.2606 


9.2680 


■/i- 


0.7320 


9.9927 


30 


40 


9.2674 


9-2750 


70 


0.7250 


9 9924 


20 


50 


9.2740 




9.2819 


Cy 


O.7181 


9.9922 


10 


U 


9.2806 


64 
64 
63 


9.2887 


66 
67 
65 
64 
63 
63 
61 
61 


O.7II3 


9.9919 


79 


10 


9.2870 


9-2953 


0.7047 


9.9917 


50 


20 


g.2934 


9.3020 


0.6980 


9.9914 


40 


30 


9 2997 


9-3085 


0.6915 


9.9912 


30 


40 


9 3058 


61 


9-3149 


0.6851 


9.9909 


20 


50 


9-3119 


60 

59 
58 


9.3212 


0.6788 


9-9907 


10 


12 

10 


9-3179 


9-3275 


0.6725 


9.9904 


78 


9-3238 


9-3336 


0.6664 


9.9901 


50 


20 


9-3296 


9-3397 




0.6603 


9.9899 


40 


30 


9-3353 




9-3458 




0.6542 


9.9896 


30 


40 


9.3410 


56 

55 


9-3517 




0.6483 


9-9893 


20 


50 


9.3466 


9-3576 


58 


0.6424 


9.9890 


10 


13 


9-3521 


9.3634 


0.6366 


9.9887 


77 




L. Cos. 


DifC. 


L. Cotg. Diff. 


L. Tang. 


L. Sin. 


/ 



Prop. Parts. 




138 127 117 


14. 13. 12. 


28. 25. 23 




41- 38. 35 




55- 51. 47 




69. 64. 59 




83. 76. 70 




97. 89. 82 




no. 102. 94 




124. 114. 105. 


108 101 94 


II. 10. 9.4 


22 


20. 18.8 


32 


30. 28.2 


43 


40. 37-6 


54 


50. 47.0 


65 


61. 56.4 


76 


71- 65.8 


86 


81. 75-2 


97 


91. 86.4 


89 84 80 


8,-9 8.4 8.0 


i7.~8 16.8 16.0 


26.7 25.2 24.0 


35-6 33-6 32-0 


44.5 42.0 40.0 


53.4 50.4 48.0 


62.3 58.8 56.0 


71.2 67.2 64.0 


80.1 75.6 72.0 


76 73 69 


7.6 7.3 6.9 


15.2 14.6 13.8 


22.8 21.9 20.7 


30.4 29,2 27.6 


38.0 36.5 34.5 


45.6 43.8 41.4 


53.2 51. I 48.3 


60.8 58.4 55-2 


68.4 65.7 62.1 


67 63 61 


6.7 6.3 6.1 


13.4 12.6 12.2 


20.1 18.9 18.3 


26.8 25.2 24.4 


33-5 31-5 30.5 


40.2 37.8 36.6 


46.9 44.1 42.7 


53.6 50.4 48.8 


60.3 56.7 54.9 


59 57 55 


5-9 5-7 5-5 


II. 8 II. 4 II. 


17.7 17. I 16.5 


23.6 22.8 22.0 


29.5 28.5 27.5 


35.4 34.2 33.0 


41.3 39.9 38.5 


47.2 45.6 44-0 


53 


I 51-3 49-5 



77° - 85°. 



178 









13^ 


— 


21°. 






1 L. Sin. Diff' 


L.Tang. lDiff| 


L. Cotgf. 


L. Cos. 




13°0'i 9.3521 


54 


9-3634 


57 


0.6366 


9.9887 


0' 77^ 


10 9-3575 


9.3691 


0.6309 


9.9884 


50 


20 9.3629 


b4 


9-3748 


56 


0.6252 


9.9881 


40 


30 


9.3682 


bi 


9-3804 


0.6196 


9.9878 


30 


40 


9-3734 


b'^ 


9-3859 


55 


O.6141 


9-9875 


20 


50 


9-3786 


b^ 
51 
50 


9-3914 
9-3968 


55 
54 
53 


0.6086 


9.9872 


10 


M 1 9.3837 


0.6032 


9.9869 76 1 


10 


9.3887 


9.4021 


0.5979 


9.9866 


50 


20 


9-3937 


5° 


9.4074 


53 


0.5926 


9-9863 


40 


30 


9-3986 


49 


9.4127 




0.5873 


9.9859 


30 


40 


9-4035 


49 
48 

47 
47 
46 
46 


9.4178 




0.5822 


9-9856 


20 


50 


9-4083 


9-4230 


52 

51 

50. 

50 

49 

49 

48 

48 

47 

47 


0.5770 


9-9853 


10 


15 


9.4130 


9.4281 


0.5719 


9.9849 


75 


10 


9-4177 


9-4331 


0.5669 


9.9846 


50 


20 


9-4223 


9-4381 


0.5619 


9-9843 


40 


30 


9.4269 


9.4430 


0.5570 


9.9839 


30 


40 


9-4314 




9.4479 


0.5521 


9-9836 


20 


50 


9-4359 


44 
44 

44 


9-4527 


0.5473 


9.9832 


10 


16 


9.4403 


9-4575 


0.5425 


9.9828 


74 


10 


9-4447 


9.4622 


0.5378 


9.9825 


50 


20 


9.4491 


9.4669 


0.5331 


9.9821 


40 


30 


9-4533 




9.4716 


46 
46 
45 
45 
45 


0.5284 


9.9817 


30 


40 


9-4576 




9-4762 


0.5238 


9.9814 


20 


50 


9.4618 


41 
41 
41 


9.4808 


0.5192 


9.9810 


10 


17 


9.4659 


9-4853 


0.5147 


9 . 9806 


73 


10 


9.4700 


9.4898 


0.5102 


9 . 9802 


50 


20 


9.4741 


40 
40 


9-4943 


44 
44 
44 
43 
43 


0.5057 


9.9798 


40 


30 


9-4781 


9.4987 


0.5013 


9.9794 


30 


40 


9.4821 


9-5031 


0.4969 


9.9790 


20 


50 


9.4861 


39 
39 


9-5075 


0.4925 


9.9786 


10 


18 


9.4900 


9-5118 


0.4882 


9-9782 


72 


10 


9-4939 


9-5161 


0.4839 


9.9778 


50 


20 


9-4977 


78 


9-5203 




0.4797 


9-9774 


40 


30 


9-5015 


37 
38 
36 
37 

^6 


9-5245 


42 
42 
41 
41 
40 


0.4755 


9.9770 


30 


40 


9-5052 


9-5287 


0.4713 


9-9765 


20 


50 


9-5090 


9-5329 


0.4671 


9.9761 


10 


19 


9-5126 


9-5370 


0.4630 


9-9757 


71 


10 


9-5163 


9-54II 


0.4589 


9-9752 


50 


20 


9-5199 


^fi 


9-5451 


40 


0.4549 


9-9748 


40 


30 


9-5235 




9-5491 




0.4509 


9-9743 


30 


40 


9.5270 


36 


9-5531 


40 


0.4469 


9-9739 


20 


50 


9-5306 


35 
34 
34 


9-5571 


40 

39 
39 


0.4429 


9-9734 


10 


20 


9-5341 


9-5611 


0.4389 


9.9730 


70 


10 


9-5375 


9-5650 


0.4350 


9-9725 


50 


20 


9.5409 


34 


9.5689 


q8 


O.4311 


9.9721 


40 


30 


9-5443 




9-5727 




0.4273 


9.9716 


30 


40 


9-5477 




9.5766 


38 
38 


0.4234 


9.9711 


20 


50 1 9.5510 


33 


9-5804 


0.4196 


9.9706 


10 


21 19-5543 


9-5842 


0.4158 


9.9702 


69 


1 L. Cos. 


Diff 


L. Cotg. 


Diff 


L. Tang. 


L. Sin. 





Prop. 


Parts. 


67 


55 


53 


5-7 


5-5 


5-3 


II. 4 


II. 


10.6 


17. 1 


16.5 


15-9 


22.8 


22.0 


21 2 


28.5 27.5 


26.5 


34 2 


33-0 


31.8 


39-9 


38.5 


37-1 


45-0 


44 -o 


42.4 


51.3 49.5 47.7 


52 


50 


48 


5-2 


5-0 


4.8 


10.4 


10. 


9.6 


15.6 


I5-0 


14.4 


20 8 


20.0 


19.2 


26.0 


2S.O 


24.0 


31.2 


30.0 


28.8 


36-4 35-0 


33-6 


41.6 


40.0 


38.4 


46.8 


45-0 


43-2 


46 


44 


42 


4.6 


4.4 


4.2 


Q.2 


8.8 


8.4 


13.8 


13.2 


12.6 


18.4 


17.6 


16.8 


23.0 


22.0 


21.0 


27.6 26.4 25.2 


32.2 


30.8 


29.4 


36.8 


35-2 


33-6 


41.4 


39-0 


37-8 


41 


40 


39 


4.1 


4.0 


3.9 


8.2 


8.0 


7.8 


12.3 


12.0 


II. 7 


16.4 


16.0 


15-6 


20.5 


20.0 


19-5 


24.0 


24.0 


23.4 


28.7 


28.0 


27-3 


32.8 


32.0 


31.2 


36.9 


36.0 


35.1 


38 


37 


36 


3.8 


3-7 


3-6 


7.6 


7-4 


7.2 


II. 4 


II. I 


10.8 


15-2 


14.8 


14.4 


19.0 


18. s 


18.0 


22.8 


22.2 


21.6 


26.6 


2S.Q 


25.2 


30.4 29.6 28.8 


34-2 


33-3 32.4 


35 


34 


33 


3.5 


3-4 


3 3 


7.0 


6.8 


6.6 


10. s 


10.2 


9.9 


14.0 


13.0 


13.2 


17.5 17.0 


16.5 


21.0 


20.4 


19.8 


24-5 


23.8 


23.1 


28.0 


27.2 


26.4 


31-5 


30.6 


29.7 



69^ 



77°. 



179 









2r 


— 


29°. 








L. Sin. 


DifE 


L. Tang. 


Diff 


L. Cotg. 


L. Cos. 




21° 0' 


9-5543 


33 


9.5842 


37 
38 


0.4158 


9.9702 


0' 69° 


10 


9-5576 


9-5879 


O.4121 


9-9697 


50 


20 


9-5609 


33 


9-5917 


0.4083 


9.9692 


40 


30 


9-5641 


3-! 


9-5954 


3/ 


. 4046 


9.9687 


30 


40 


9-5673 


3^ 


9-5991 


3; 


0.4009 


9.9682 


20 


50 


9-5704 


3^ 
32 
31 


9.6028 


61 
36 
36 
36 
36 
36 


0.3972 


9-9677 


10 


22 


9-5736 


9.6064 


0.3936 


9.9672 


68 


10 


9-5767 


g.6ioo 


0.3900 


9.9667 


50 


20 


9-5798 


3^ 


9.6136 


0.3864 


9.9661 


40 


30 


9.5828 


3^ 


9.6172 


0.3828 


9-9656 


30 


40 


9-5859 


3i 


9.6208 


0.3792 


9.9651 


20 


50 
23 


9.5889 


3" 
30 
29 


9-6243 


3i 
36 
35 


0.3757 


9.9646 


10 


9-5919 


9.6279 


0.3721 


9.9640 


67 


10 


9-5948 


9.6314 


0.3686 


9-9635 


50 


20 


9-5978 


3'^ 


9.6348 




0.3652 


9.9629 


40 


30 


9 . 6007 


2y 


9-6383 


35 


0.3617 


9.9624 


30 


40 


9.6036 


2y 


9.6417 


34 


0.3583 


9.9618 


20 


50 


9.6065 


28 
28 

08 


9.6452 


35 

3: 


0.3548 


9.9613 


10 


24 


9.6093 


9.6486 


0.3514 


9.9607 


66 


10 


9.6121 


9.6520 


0.3480 


9 . 9602 


50 


20 


9.6149 


„o 


9-6553 




0.3447 


9-9596 


40 


30 


9.6177 


„o 


9.6587 




0.3413 


9.9590 


30 


40 


9.6205 




9.6620 




0.3380 


9-9584 


20 


50 


9.6232 


27 
27 


9-6654 


34 
33 
33 
32 
33 


0.3346 


9-9579 


10 


25 


9.6259 


9.6687 


0.3313 


9-9573 


65 


10 


9.6286 


9.6720 


0.3280 


9-9567 


50 


20 


9-6313 




9.6752 


0.3248 


9.9561 


40 


30 


9-6340 




9.6785 


0.3215 


9-9555 


30 


40 


9.6366 




9.6817 




0.3183 


9-9549 


20 


50 


9.6392 


26 
26 

"6 


9.6850 


32 
32 


0.3150 


9-9543 


10 


26 


9.6418 


9.6882 


O.3118 


9-9537 


64 


10 


9-6444 


9.6914 


0.3086 


9-9530 


50 


20 


9.6470 


25 

•-6 


9.6946 


31 
32 


0.3054 


9-9524 


40 


30 


9-6495 


9.6977 


0.3023 


9.9518 


30 


40 


9-6521 




9.7009 


0.2991 


9.9512 


20 


50 


9-6546 


24 
25 


9.7040 


32 
31 


0.2960 


9.9505 


10 


27 


9-6570 


9.7072 


0.2928 


9.9499 


63 


10 


9-6595 


9.7103 


0.2897 


9.9492 


50 


20 


9.6620 


24 
24 
24 
24 
24 


9-7134 


31 
31 
30 
31 
30 


0.2866 


9.9486 


40 


30 


9.6644 


9.7165 


0.2835 


9-9479 


30 


40 


9.6668 


9.7196 


0.2804 


9-9473 


20 


50 


9 . 6692 


9.7226 


0.2774 


9.9466 


10 


28 


9.6716 


9-7257 


0.2743 


9-9459 


62 


10 


9.6740 


9.7287 


0.2713 


9-9453 


50 


20 


9.6763 




9-7317 




0.2683 


9.9446 


40 


30 


9.6787 


23 


9-7348 


■30 


0.2652 


9-9439 


30 


40 


9.6810 


9-7378 


30 
30 


0.2622 


9.9432 


20 


50 


9-6833 


9.7408 
9-7438 


0.2592 


9-9425 


10 


29 


9.6856 


0.2562 


9.9418 


61 




L. Cos. 


DifEl L. Cotg. iDifE 


L. Tang, i L. Sin. i | 



Prop. Parts. 





37 86 35 


I 

2 

3 


3-7 3-6 3-5 
7.4 7.2 7.0 
II. I 10.8 10.5 


4 

% 


14.8 14.4 14.0 
18.5 18.0 17.5 
22.2 21.6 21.0 


7 
8 


25.9 25.2 24.5 
29.6 28.8 28.0 


9 


33-3 32.4 31-5 



34 33 



3-4 
6.8 



3-3 
6.6 



3-2 
6.4 
9.6 



10.2 9.9 
13.6 13.2 12.8 
17.0 16.5 16.0 
20.4 19.8 19.2 
23.8 23.1 22.4 
27.2 26.4 25.6 
30.6 29.7 28.8 



31 

3-1 



30 

3-0 
6.0 



9.0 
12.0 II. 6 



29 

2.9 
5-8 
"7 



6.2 
9-3 
12.4 

15.5 15.0 14. 5 

18.6 18.0 17.4 

21.7 21.0 20.3 

24.8 24.0 23.2 

27.9 27.0 26.1 



26 

2.6 

5-2 

7.8 

10.4 
I3-0 
iS-6 
18.2 
22.4 21.6 20.8 
25.2 24.3 23.4 



28 27 

2.8 2.7 

S-6 5-4 

8.4 8.1 

II. 2 10.8 

14-0 13-5 

16.8 16.2 

19.6 18.9 



24 28 

2.4 2.3 

4.8 4.6 

7.2 6.9 

9.6 9.2 

12.0 II. 5 

14.4 13.8 

16.8 16. I 

20.0 19.2 18.4 

22.5 21.6 20.7 



25 

2.5 
S-o 
7-5 
10. o 
12. s 
150 
17-5 



61° — 69 



180 









29 


3 


37. 








1 L. Sin. 


Diff L. Tang". 


Difif L. Cotg-. 


L. Cos. 1 1 


29= O'l 9.6856 


22 


9-7438 


29 


0.2562 


9.9418 


0'61° 


10 ! 9.6878 


9.7467 


0.2533 


9 


.9411 


50 


20 


9.6901 


^S 


9-7497 


J^ 


0.2503 


9 


9404 


40 


30 


9.6923 




9-7526 


29 


0.2474 


9 


-9397 


30 


40 


9.6946 




9-7556 


3'-J 


0.2444 


9 


-9390 


20 


50 


9.6968 


22 
22 


9-7585 


2y 
29 
30 


0.2415 


9 


-9383 


10 


30 


9.6990 


9-7614 


0.2386 


9 


-9375 


60 


10 


9.7012 


9.7644 


0.2356 


9 


9368 


50 


20 


9-7033 




9.7673 




0.2327 


9 


9361 


40 


30 


9-7055 




9-7701 




0.2299 


9 


9353 


30 


40 


9.7076 




9 -7730 


29 


0.2270 


9 


9346 


20 


50 


9.7097 


21 
21 


9-7759 


29 
29 
28 


0.2241 


9 


9338 


10 


31 i 9.711S 


9-7738 


0.2212 


9 


9331 


59 


10 


9-7139 


9.7816 


0.2184 


9 


9323 


50 


20 


9.7160 




9-7845 




0.2155 


9 


9315 


40 


30 


9.7181 




9-7873 




0.2127 


9 


9308 


30 


40 


9.7201 




9.7902 




0.2098 


9 


9300 


20 


50 


9.7222 


20 
20 


9-7930 


28 
28 


0.2070 9 


9292 


10 


32 


9.7242 


9-7958 


. 2042 


9 


9284 


58 


10 


9.7262 


9.7986 


0.2014 


9 


9276 


50 


20 


9.7282 




9.8014 


28 


0.1986 


9 


9268 


40 


30 


9-7302 




9.8042 


0. 


0.1958 


9 


9260 


30 


40 


9-7322 




9 . 8070 




0.1930 


9 


9252 


20 


50 


9-7342 


19 
19 


9.8097 


28 
28 
27 
28 


0.1903 


9 


9244 


10 


33 9.7361 


9-8125 


0.1875 


9 


9236 


57 


10 


9.7380 


9-8153 


0.1847 


9 


9228 


50 


20 


9.7400 


19 


9.8180 


0.1820 


9 


9219 


40 


30 


9.7419 


9.8208 




0.1792 


9 


9211 


30 


40 19-7438 




9 8235 


28 


0.1765 


9 


9203 


20 


50 19-7457 


19 
18 

19 
18 


9-8263 


27 
27 
27 
27 
27 
27 
27 
27 

27 


0.1737 


9 


9194 


10 


U 19-7476 


9.8290 


0.1710 


9 


9186 


56 


10 1 9.7494 


9-8317 


0.1683 


9 


9177 


50 


20 9-7513 


9-8344 


1656 


9 


9169 


40 


30 9-7531 


19 
18 


9-8371 


0.1629 


9 


9160 


30 


40 9-7550 


9-8398 


0.1602 


9 


9151 


20 


50 1 9.7568 


18 
18 
t8 


9.8425 


0.1575 


9 


9142 


10 


35 \ 9-7586 


9.8452 


0.1548 


9 


9134 


55 


10 9.7604 


9.8479 


0.1521 


9 


9125 


50 


20 1 9.7622 


18 


9.8506 


27 

"6 


0.1494 


9 


9116 


40 


30 9.7640 


17 

t8 


9-8533 


0.1467 


9 


9107 


30 


40 


9-7657 


9-8559 


27 
27 
26 


0.1441 


9 


9098 


20 


50 


9-7675 


17 

18 


9.8586 


0.1414 


9 


9089 


10 


36 


9.7692 


9.8613 


0.1387 


9 


9080 


54 


10 


9.7710 


9-8639 


0.1361 


9 


9070 


50 


20 


9.7727 


17 


9.8666 


76 


0.1334 


9 


9061 


40 


3019.7744 


17 


9.8692 


06 


0.1308 


9 


9052 


30 


40 ! 9.7761 


17 


9.8718 


27 


0.1282 


9 


9042 


20 


50 1 9.7778 


17 


9-8745 


26 


0.1255 


9 


9033 


10 - 


37 0:9.7795 


9.8771 


0.1229 


9 


9023 


53 


1 L. Cos. IDifE 


L. Cotg. 


Diff 


L. Tang. 


L. Sin. 





Prop. 


Parts. 


29 


28 27 


2.9 


2.8 2.7 


^.8 


5-6 5-4 


8.7 


8.4 8.1 


II. 6 


II. 2 10.8 


14-5 


14.0 13.5 


17.4 


16.8 16.2 


20.3 


19.6 18.9 


23.2 


22.4 21 .6 


26.1 


25-2 24.3 


26 


25 24 


2.6 


2.5 2.4 


5-2 


5.0 4-8 


7.8 


7-5 7-2 


10.4 


10. 9.0 


13.0 


12.5 12.0 


iS.b 


15.0 14.4 


18.2 


17.5 16.8 


20.8 


20.0 19.2 


23-4 


22.5 21.6 


23 


22 21 


2.3 


2.2 2.1 


4.6 


4.4 4.2 


6.Q 


6.6 6.3 


9.2 


8.8 8.4 


"•5 


II. ID. 5 


13.8 


13.2 12.6 


16. 1 


15.4 14.7 


18.4 


17.6 16.8 


20.7 


19.8 18.9 


20 


19 


2.0 


1.9 


4.0 


3-8 


6.0 


5-7 


8.0 


7.6 


10. 


9-5 


12.0 


II. 4 


14.0 


133 


16.0 


15.2 


18.0 


17. 1 


18 


17 


1.8 


1-7 


3-6 


3-4 


5-4 


5-1 


7.2 


6.8 


9.0 


8.5 


10.8 


10.2 


12.6 


II. 9 


14.4 


13-6 


16.2 


15.3 



53° — 61^ 



181 











37' 


— 


45% 








L. Sin. ] 


Diff 


L. Tang. 


Diff 


L. Cotg. 


L. Cos. 




37° 0' 


9-7795 


16 


9.8771 


26 


0.1229 


9-9023 


0'53° 


10 


9 


7811 


9.8797 


0.1203 


9.9014 


50 


20 


9 


7828 


16 


9.8824 


'^1 
26 


O.II76 


9.9004 


40 


30 


9 


7844 


9.8850 


O.I150 


9.8995 


30 


40 


9 


7861 


^1 
16 

16 
17 


9.8876 




O.I124 


9-8985 


20 


50 


9 


7877 


9 . 8902 


26 
26 


0.1098 


9-8975 


10 


38 


9 


7893 


9.8928 


0.1072 


9-8965 


52 


10 


9 


7910 


9.8954 


0.1046 


9-8955 


50 


20 


9 


7926 




9.8980 




0.1020 


9-8945 


40 


30 


9 


•7941 




9 . 9006 


26 
26 
26 
26 


0.0994 


9-8935 


30 


40 


9 


•7957 




9 •9032 


0.0968 


9.8925 


20 


50 


9 


•7973 


16 
15 


9.9058 


. 0942 


9.8915 


10 


39 


9 


7989 


9.9084 


0.0916 


9-8905 


51 


10 


9 


8004 


9.9110 


0.0890 


9-8895 


50 


20 


9 


8020 




9^9135 




0.0865 


9.8884 


40 


30 


9 


8035 




9.9161 


26 


0.0839 


9 8874 


30 


40 


9 


.8050 


16 
IS 

15 


9.9187 


0.0813 


9.8864 


20 


50 


9 


8066 


9.9212 


26 
26 


0.0788 


9-8853 


10 


40 


9 


8081 


9^9238 


0.0762 


9,8843 


50 


10 


9 


8096 


9.9264 


0.0736 


9.8832 


50 


20 


9 


8111 




9.9289 


26 


O.0711 


9.8821 


40 


30 


9 


8125 




9 9315 




0685 


9.8810 


30 


40 


9 


8140 




9.9341 




0.0659 


9.8800 


20 


50 


9 


8155 


14 
15 
14 


9.9366 


26 

25 

06 


0,0634 


9.8789 


10 


41 


9 


8169 


9 9392 


. 0608 


9 8778 


49 


10 


9 


8184 


9.9417 


0.0583 


9.8767 


50 


20 


9 


8198 


9-9443 


25 
26 


0.0557 


9.8756 


40 


30 


9 


8213 




9.9468 


0.0532 


9-8745 


30 


40 


9 


8227 




9.9494 




0506 


9-8733 


20 


50 


9 


8241 


14 
14 
14 
14 


9-9519 


25 
26 

IK 


0.0481 


9.8722 


10 


42 


9 


8255 


9-9544 


0.0456 


9.8711 


48 


10 


9 


8269 


9-9570 


0.0430 


9 8699 


50 


20 


9 


8283 


9-9595 


26 


0.0405 


9 8688 


40 


30 


9 


8297 


9.9621 


25 


0.0379 


9 8676 


30 


40 


9 


8311 




9 9646 


0.0354 


9.8665 


20 


50 


9 


8324 


14 

13 


9.9671 


26 
25 


0.0329 


9-8653 


10 


43 


9 


8338 


9,9697 


0.0303 


9.8641 


47 


10 


9 


8351 


9.9722 


0.0278 


9 8629 


50 


20 


9 


8365 


13 
13 
14 
13 
13 
13 


9.9747 


25 
26 


0.0253 


9 8618 


40 


30 


9 


8378 


9.9772 


0.0228 


9 8606 


30 


40 


9 


8391 


9.9798 


25 
25 
26 
25 


0.0202 


9 8594 


20 


50 


9 


8405 


9-9823 


0.0177 


9.8582 


10 


44 


9 


8418 


9 9848 


0.0152 


9 8569 


46 


10 


9 


8431 


9.9874 


0.0126 


9 8557 


50 


; 20 


9 


8444 




9.9899 




O.OIOI 


9 8545 


40 


30 


9 


8457 




9.9924 




0.0076 


9 8532 


30 


40 


9 


8469 




9-9949 


"6 


0.0051 


9.8520 


20 


50 


9 


8482 


13 


9-9975 
0.0000 


25 


0.0025 


9-8507 


10 


45 


9 


8495 


0.0000 


9.8495 


45 


, 


L. Cos. !] 


Diff 


L. Cotg. 


Diff 


L. Tang. 


L. Sin, 





Prop. Parts. 

26 25 24 

2.6 2.5 2.4 

S-2 5-0 4-8 

7-8 7-5 7-2 

10.4 10. o 9.6 

13.0 12.5 12.0 

15.6 15.0 14.4 

18.2 17.5 16.8 

20.8 20.0 19.2 

23.4 22.5 21.6 



22 21 



2.3 2 

4'6 4 
6.9 6 
9.2 8 
II. 5 II 
13-8 13 
16. I 15 
18.4 17 
20.7 19 



2.1 
4.2 

6.3 
8.4 
10.5 
12.6 
14.7 
16.8 



20 19 18 

2.0 1.9 1.8 

4.0 3.8 3.6 

6.0 5.7 5.4 

8.0 7.6 7.2 

10. o 9.5 9.0 

12.0 II. 4 10.8 

14.0 13.3 12.6 

16.0 15.2 14.4- 

18.0 17.1 16.2 



17 

1-7 

3 
5 
6 



16 

1.6 
3-2 

4.8 
6.4 
8.0 
9.6 

II. 2 
12.8 
14.4 



15 

1-5 
3-0 
4-5 
6.0 

7-5 
9.0 
to. 5 



14 13 12 

1.4 1.3 1.2 

2.8 a. 6 2.4 

4.2 3.9 3.6 

5-6 5-2 4-8 

7.0 6.5 6.0 

8.4 7.8 7.2 

9.8 9.1 8.4 

II. 2 10.4 9.6 

12.6 11.7 10.8 



45° — 53^ 



182 



Theee-plaoe Tables. 



[183] 







Table of Thbee-Place Logarithms. 






1 


.ooo 


301 


51 


708 


8 
8 


101 


.C04 




151 


.179 




2 


.3or 


176 


52 


716 


102 


.009 


4 


152 


.182 





8 


• 477 


125 


53 


724 


8 


103 


.013 


4 


158 


.185 


3 


4 


.602 


97 
79 
67 


54 


732 


8 


104 


.017 


4 
4 
4 
4 
4 
4 
4 


154 


.188 




5 


.699 


55 


740 


8 


105 


.021 


155 


.190 


3 

3 

3 


6 


.778 


56 . 


748 


8 


106 


.025 


156 


.193 


7 


.845 


58 
51 
46 
41 


57 . 


756 


7 
8 


107 


.029 


157 


.196 


8 


.903 


58 . 


763 


108 


•033 


158 


.199 


9 


.954 


59 


771 




109 


.037 


159 


.201 


3 
3 


10 


.000 


60 


778 


110 


.041 


160 


.204 


11 


.041 


38 


61 


785 




111 


.045 


4 


161 


.207 


3 


12 


.079 




62 


792 




112 


.049 




162 


.210 




IB 


.114 


32 

30 
28 


63 


799 


7 


113 


•053 


4 
4 


163 


.212 


3 


14 


.146 


64 


806 


114 


•057 


164 


.215 


15 


.176 


65 


813 


115 


.061 


165 


.217 




16 


.204 


26 


66 


820 


6 


116 


.064 




166 


.220 




17 


.230 




67 


826 




117 


.068 




167 


.223 




18 


.255 




68 


833 


5 


118 


.072 




168 


.225 




19 


•279 


22 
21 


69 


839 


6 
6 
6 


119 


.076 


3 
4 


169 


.228 


2 
3 


20 


.301 


70 


845 


120 


.079 


170 


.230 


21 


.322 


71 


851 


121 


.083 


171 


•233 


22 


• 342 


20 


72 


8S7 


6 


122 


.086 


4 


172 


.236 


2 


28 


.362 


18 


73 


863 


6 


123 


.090 





178 


.238 


3 


24 


.380 


18 


74 


869 


6 


124 


.093 




174 


.241 




25 


.39B 




75 


875 


6 


125 


.097 


3 


175 


.243 


3 


26 


• 415 


16 
16 


76 


881 




126 


.100 


176 


.246 


27 


• 431 


77 


886 


6 


127 


.104 




177 


.248 




28 


-447 




78 


892 


6 
5 

5 


128 


.107 




178 


.250 




29 


.462 


15 

15 
14 


79 


898 


129 


.III 


4 
3 
3 


179 


.253 


2 
3 


30 


.477 


80 


903 


130 


.114 


180 


.255 


31 


.491 


81 


908 


181 


.117 


181 


.2^8 


32 


•505 


82 


914 




182 


.121 




182 


.260 




33 


•519 




88 


919 


5 


188 


.124 


3 


183 


.262 


3 


34 


.531 




84 


924 




134 


.127 




184 


.26s 


35 


.544 




85 


929 




135 


.130 


3 


185 


.267 




36 


.556 




86 


934 


6 


136 


•134 




186 


.270 


3 


37 


.568 




87 


940 


137 


.137 




187 


.272 




38 


.580 




88 


944 




138 


.140 


3 


188 


.274 




39 


•591 


II 
II 


89 


949 


5 
5 
5 


189 


• 143 


3 
3 
3 
3 


189 


.276 


3 
2 


40 


.602 


90 


954 


140 


.146 


190 


.279 


41 


.613 


91 


959 


141 


.149 


191 


.281 


42 


.623 




92 


964 




142 


.152 


192 


.283 




43 


.633 




93 


968 




143 


-155 




193 


.286 




44 


•643 




94 


973 




144 


.158 




194 


.288 




45 


.6^3 




95 


978 




145 


.161 




195 


.290 


" 


46 


.663 




96 


982 


4 


146 


.164 


3 


196 


.292 


2 


47 


.672 




97 


987 




147 


.167 




197 


•294 


3 


48 


.681 




98 


991 




148 


.170 




198 


•297 


49 


.690 


9 
9 


99 


996 


5 
4 


149 


.173 


3 

3 


199 


.299 


2 


50 


.699 


100 


000 


150 


.176 


200 


.301 



184 







Natural Trigonometrical Functions. 








Sin. 


^ Tan. 




Sec. 


SC 


Cosec. 


id 


Cot, 


■*i 


-|§ 


-1 






Q 


_ 
17 

t8 









Q 




5 




0° 


.ooo 


.000 
18 •017 


1. 000 










I. 000 ^ 


90° 


1 


.017 


1. 000 


T 


57- 299 


28.6 


57-290 


28.6 


I . 000 J 


89 


3 

4 


•035 

.052 
.070 


17 -^35 

x8 •°52 

[I -087 


17 
18 

7s 


I 001 
1. 001 
1.002 




I 



28 654 
19,107 
14.336 


9.94 

4-77 
2.86 


28,636 
19.081 
14.301 


9-55 
4-78 
2.87 


•999 
.999 I 
.998 , 


88 
87 
86 


5 


.087 


1.004 




II 474 




II 430 


.996 ^ 


8b 


6 


-105 


17 -^^s 

17 -123 

17 -141 

8 -^58 


t8 


1.006 




9-567 


1.36 
1.02 


9-514 


1-37 

1.03 

.801 

-643 

.526 


995 2 


84 


7 
8 


.122 
.139 


18 


1.008 
1. 010 


2 


8.206 
7-185 


8.144 
7. 115 


•993 _ 
•990 I 


83 

82 


9 


.10 


iS 
18 


1. 012 




6.392 


.633 
.518 


6.314 


.988 I 


81 


10 


.174 


;: -^76 


I 015 


i 


5^759 


5-671 


.985 I 


80 


11 


.191 


17 -^94 


1. 019 




5.241 


5-145 


•982 I 


79 


12 


.208 


7 -213 


18 
.8 


1.022 


3 


4.810 


•431 
365 


4-705 


.440 


978 I 


78 


13 


.225 


•231 


1.026 


4 


4.445 


4-331 


-374 


-974 I 


77 


14 


.242 


•249 


1. 031 




4- 134 




4. on 




.970 ^ 


76 


15 


.259 


'7 .268 


'" 


1-035 


4 
5 


3.864 


.270 
.236 
.208 
.184 
.164 
.148 
•134 


3 732 


•279 


.966 ^ 


75 


16 


.276 


z .287 

; •3°6 


iy 


1.040 


3.628 


3-487 


.245 
.216 


.961 
956 I 
•951 ' 


74 


17 


.292 


iy 


1.046 


5 


3 420 


3.271 


73 


18 


.309 


'' .325 


iy 


1. 051 


3.236 


3.078 


-193 


72 


19 


.326 


I -344 
'' .364 
''1 .384 


iy 


1.058 


g 


3.072 


2.904 


.174 


•946 I 


71 


20 


.342 


20 


1.064 


7 
8 


2.924 


_2^747 
2.605 


-157 
.142 


940 g 


70 


21 


.3=^8 


1. 071 


2.790 


•934 ^ 


69 


22 


•375 


'I -404 


20 


1.079 


2.669 




2-475 


130 


•927 
•921 


68 


23 


.391 


1 -424 


20 


1.086 




2.559 




2.356 


.119 


67 


24 


.407 


1 -445 


21 


1.095 


8 


2.459 


..0. 


2.246 




•914 I 
•906 , 


66 


25 


•423 


^6 .466 


21 


1. 103 


2,366 


•093 
.085 
.078 


2 145 




65 


26 

27 


•438 
•454 




22 


1. 113 
1,122 


9 


2.281 
2.203 


2.050 
1.963 


.087 
.082 


•899 s 
.891 g 


64 
63 


28 


.469 


i -532 




i^i33 




2.130 


.067 
.063 
.058 
.055 


I. 881 


.883 8 
•875 ' 


62 


29 


• 485 


^'i .554 


22 


I -143 


ID 


2.063 


1.804 


.077 


61 


30 


.500 


;^! -577 


^3 
24 


I.I55 


12 


2.000 


1.732 
1.664 


.072 
.068 
.064 
.060 


.866 I 


60 


31 


•515 , 


''1 .601 


1. 167 


1.942 


•857 
,848 I 

•839 XO 


59 


32 


.530 


11 -625 




1. 179 


13 
14 
15 
15 

t6 


1.887 


1.600 


58 


33 


•545 1 


! -649 


24 
26 


1. 192 


I 836 


.048 


1.540 


57 


34 


•559 ! 


t -^75 


1.206 


1.788 


1.483 




•829 10 


56 


35 


.574 ! 


^5 .700 


^0 


1,221 


1.743 




1.428 




•819 ,. 


55 


36 


• 588 1 


!! -7^7 


^7 


1,236 


1. 701 




1.376 


.052 


•809 lio 


54 


37 


.602 


^^ ^754 


27 


1.252 




1.662 


.038 


1.327 




•799 TT 


53 


38 


.616 


;t -781 


27 


1.269 


18 

18 


I 624 


1.280 


.047 


•788 ;; 


52 


39 


.629 


xU839_ 

. -869 
I J -900 


29 


1.287 


I 589 


■uib 


1.235 


■045 


•777 „ 


51 


40 


.643 


29 
30 


1.305 


1-556 


.033 


1. 192 


-043 
.042 


.766 " 


50 
49 


41 


.656 i 


1.325 


^J 


1.524 


.032 


1-150 


•755 ^, 


42 


.669 


3i 


1.346 




1.494 


.028 
.026 
.026 


I. Ill 




743 ,, 


48 


43 


.682 


I ^933 




1.367 




1,466 


1.072 


-039 
.036 
.036 


•731 ., 


47 


44 
45 


• 695 
.707 


III -966 
i I . 000 


34 


1.390 

1-414 


-!3 


1.440 
1. 414 


1.036 
1. 000 


•719 z 
•707 


46 
45 




Cos. 


i Cot. 


d 


Cosec. 


s« 


Sec. 


!S 


Tan. 


tJ 


Sin. !« 








q 


u 




p 




Q 




Q 


Q 





185 



Logarithmic Sines, 


ETC., FOR EVERY TeNTH OF DEGREE. 


o 


' Sin. 


Diff. 


Tan. 


DifE. 


Cot. 


Cos. 





0.0 












0.000 


90.0 


0.1 


7.242 


301 
176 
125 
97 
79 

58 


7.242 




2.758 


0.000 


89.9 


0.2 


7 


•543 


7-543 


176 


2.457 


0.000 


89.8 


0.8 


7 


71Q 


7.719 


2.281 


0.000 


89.7 


0.4 


7 


.844 


7.844 


97 
79 
67 
58 


2.156 


0.000 


89.6 


0.5 


7 


.941 


7 941 


2.059 


0.000 


89.5 


O.G 


8 


020 


8.020 


1.980 


0.000 


89.4 


0.7 


8 


087 


8.087 


I 913 


0.000 


89.3 


0.8 


8 


145 


8.145 


1-855 


0.600 


89.2 


0.9 


8 


196 


46 
38 


8.196 


51 
46 

41 
38 


I 804 


0.000 


89.1 


1.0 


8 


242 


8.242 


1.758 


0.000 


89.0 


1.1 


8 


283 


8.283 


1-717 


0.000 


88.9 


1.2 


8 


321 


8.321 


1.679 


0.000 


88.8 


1.3 


8 


3S6 


35 


8.356 


35 


1.644 


0.000 


88.7 


1.4 


8 


388 


32 


8.388 


32 


1. 612 


0.000 


88.6 


1.5 


8 


418 


30 
28 
26 


8.418 


30 
28 


1.582 


0.000 


88.5 


1.6 


8 


446 


8 446 


1-554 


0.000 


88.4 


1.7 


8 


472 


8.472 


26 


1.528 


0.000 


88.3 


1.8 


8 


4Q7 


25 


8.497 


25 


1-503 


0.000 


88.2 


1.9 


8 


■^21 


24 


8.521 


24 


1.479 


0.000 


88.1 


2.0 


8 


543 


22 
21 


8 543 


22 
21 


1.457 


0.000 


88.0 


2.1 


8 


564 


8.564 


1.436 


0.000 


87.9 


2.2 


8 


584 




8.585 


21 


1-415 


0.000 


87.8 


2.3 


8 


603 


19 


8.604 


19 


1.396 


0.000 


87.7 


2 4 


8 


622 


19 


8.622 


18 


1-378 


0.000 


87.6 


2.5 


8 


640 


17 


8.640 


18 
17 


1.360 


0.000 


87.5 


2.6 


8 


657 


8.657 


1-343 


0.000 


87.4 


2.7 


8 


673 




8 674 


• 17 


I 326 


0.000 


87.3 


2 8 


8 


68q 


16 


8.689 


IS 


1. 311 


9.999 


87.2 


2.9 


8 


704 


15 


8.705 


16 


1.295 


9.999 


87.1 


3.0 


8 


719 


15 
14 


8.719 


14 
15 


1. 281 


9.999 


87.0 


3.1 


8 


733 


8 734 


1.266 


9-999 


86.9 


3.2 


8 


747 


14 


8.747 


13 


1-253 


9.999 


86.8 


3.3 


8 


760 


13 


8.761 


14 


1.239 


9 999 


86.7 


3.4 


8 


773 


13 


8 774 


13 


1.226 


9.999 


86.6 


3.5 


8 


786 


13 
12 


8.786 


12 
13 


1.214 


9.999 


86.5 


3.6 


8 


7Q8 


8.799 


1. 201 


9.999 


86.4 


3.7 


8 


810 


12 


8 811 


12 


1. 189 


9-999 


86.3 


3.8 


8 


821 


II 


8.822 


II 


1. 178 


9.999 


86.2 


3.9 


8 


833 


12 


8.834 


12 


1. 166 


9-999 


86.1 


4 


8 


844 


II 
10 


8.845 


II 
10 


I-I55 


9.999 


86.0 


4.1 


8 


854 


8.855 


I -145 


9-999 


85.9 


4.2 


8 


86^ 


10 


8.866 




I -134 


9 999 


85.8 


4.3 


8 


875 


II 


8.876 


10 


I -124 


9.999 


85.7 


4.4 


8 


88^ 


10 


8.886 


10 


1. 114 


9 999 


85. 6 


4.5 


8 


895 


10 
9 


8.896 


10 
10 


1. 104 


9.999 


85.5 


4.6 


8 


904 


8.906 


1.094 


9 999 


85.4 


4.7 


8 


913 


9 


8.915 


9 


I 085 


9.999 


85.3 


4.8 


8 


Q23 


10 


8 924 


9 


I 076 


9.998 


85.2 


4.9 


8 


q32 


9 


8-933 


9 


1.067 


9.998 


85.1 


5.0 


8 940 


8 


8.942 


9 


I 058 


9-998 


85.0 




Cos. 


DifE. 


Cot. 


DifE 


Tan. 


Sin. 





186 





Logarithmic Sines, etc., for 


EVERY ] 
Cot. 


Degree. 




o 


Sin. 


Diff. 


Tan. 


Diff. 


Cos. 


















0.000 


90 


1 


8.242 


8.242 


1-758 


0.000 


89 


2 


8 


543 


301 


8.543 


301 


1-457 


0.000 


88 


3 


8 


719 


176 


8.719 


176 


1. 281 


9.999 


87 


4 


8 


844 


125 


8.845 


126 


I-155 


9.999 


86 


5 


8 


940 


96 


8.942 


97 


1.058 


9-998 


85 


6 


9 


019 


79 


9.022 




0.978 


9.998 


84 


7 


Q 


086 


67 


9.089 


67 


0.911 


9-997 


83 


8 


Q 


144 


58 


9.148 


59 


0.852 


9.996 


82 


9 


9 


194 


50 
46 

41 


9.200 


52 
46 
43 


0.800 


9-995 


81 


10 


9 


240 


9.246 


0.754 


9-993 


80 


11 


9 


281 


9.289 


0.711 


9-992 


79 


12 


9 


318 


37 


9-327 


38 


0.673 


9.990 


78 


13 


9 


352 


34 


9-363 


36 


0.637 


9.989 


77 


14 


9 


384 


32 
29 


9-3^:7 


34 


0.603 


9.987 


76 


15 


9 


413 


9.428 


31 


0.572 


9-985 


75 


16 


9 


440 


26 


9-457 


29 


0.543 


9-983 


74 


17 


9 


466 




9-485 


28 


0.515 


9.981 


73 


18 


9 


490 




9.512 


27 


0.488 


9-978 


72 


19 


. 9 


513 


21 
20 


9-537 


25 
24 
23 


0.463 


9-976 


71 


20 


9 


534 


9.561 


0.439 


9-973 


70 


21 


9 


554 


9-584 


0.416 


9-970 


69 


22 


9 


574 




9.606 


22 


0.394 


9-967 


68 


23 


9 


592 




9.628 


22 


0.372 


9.964 


67 


24 


9 609 


17 
17 
16 


9.649 


21 


0.351 


9.961 


66 


25 


9 626 


9.669 


20 • 


0.331 


9-957 


65 


26 


9.642 




9.688 


19 


0.312 


9-954 


64 


27 


9-657 


^^ 


9.707 


19 


0.293 


9 -950 


63 


28 


9.672 




9.726 


19 
18 

X7. 
18 


0.274 


9-946 


62 


29 


9.686 


13 


9-744 


0.256 


9.942 


61 


30 


9.699 


9.761 


0.239 


9-938 


60 


31 


9.712 




9-779 


0.221 


9-933 


59 


32 


9 724 




9.796 




0.204 


9.928 


^S 


33 


9-736 




9.813 


16 


0.187 


9.924 


57 


34 


9-748 




9.829 


0.171 


9.919 


56 


35 


9-759 




9-845 


16 
16 
16 


0.155 


9-913 


55 


36 


9.769 




9.861 


0.139 


9.908 


54 


37 


9-779 


10 


9-877 


0.123 


9.902 


53 


38 


9.789 




9-893 


0.107 


9.897 


52 


39 


9-799 


9 

9 


9.908 


15 
16 
15 


0.092 


9.891 


51 


40 


! 9.808 


9.924 


0.076 


9.884 


50 


41 


9 817 


9-939 


0.061 


9.878 


49 


42 


9.826 


8 
8 


9-954 


15 


0.046 


9.871 


48 


43 


9-834 


9.970 




0.030 


9.864 


47 


44 


9.842 


9-9S5 


15 


0.015 


9-857 


46 


45 


9.849 


7 


0.000 


15 


000 


9.849 


45 


o 

1 


Cos. 


Diff. 


Cot. 


Diff. 


Tan. 


Sin. 






187 



